THE  LIBRARY 

OF 

THE  UNIVERSITY 
OF  CALIFORNIA 

LOS  ANGELES 


GIFT  OF 

John  S.Prell 


THE 


PRINCIPLES 


ELEMENTARY  MECHANICS. 


DE    V  OLSON    WOOD, 

PROFESSOR   OF   MATHEMATICS   AND    MECHANICS   IN   THE   STEVENS    INSTITUTE 
OF  TECHNOLOGY. 


FIFTH  EDITION. 

S.  PRELL 

Civil  &  Mechanical  Engineer. 

SAN  FRANCISCO,  CAL. 

NEW  YORK: 
JOHN  WILEY  &  SONS, 


15    ASTOR    PLACE. 

1884. 


COPYRIGHT  BY 
DB    VOLSON    WOOD, 

1877. 


TROW'S 

PRINTING  AND  BOOKBINDING  Co., 

205-213  East  \-ith  St., 

NRW  YORK. 


ybrary 


PREFACE. 


THIS  work  is  especially  designed  to  treat  of  the 
pies  of  Rational  Mechanics,  and  not  to  present  a  system 
of  analysis.  The  analysis  employed  in  the  demonstration 
of  principles  is  of  an  elementary  character,  the  Calculus 
being  entirely  avoided.  A  few  problems  are  solved  which 
very  properly  belong  to  the  Calculus,  but  the  solutions 
have  been  effected  by  means  of  the  well-known  properties 
of  certain  curves  and  the  principles  of  elementary  geome- 
try. As  examples  of  this  mode  of  reasoning,  reference  is 
made  to  the  following  problems :  The  determination  of  the 
centre  of  gravity  of  a  circular  arc ;  The  time  of  vibration 
of  a  simple  pendulum ;  and  The  quantity  of  flow  of  a 
liquid  through  a  weir.  A  few  problems  are  solved  which 
involve  a  knowledge  of  Conic  Sections,  but  these  may  be 
omitted,  if  desired,  without  detriment  to  the  other  portions 
of  the  work. 

The  Articles  are  as  independent  of  each  other  as  they  can 
be  practically,  and  at  the  same  time  present  the  subject  in 
a  connected  manner.  This  feature  will  enable  the  teacher 
the  more  easily  to  select  particular  portions  of  the  work 
when  the  whole  cannot  be  taken. 

The  manner  of  applying  the  principles  of  the  subject 

733279 


IT  PREFACE. 

is  shown  by  means  of  numerous  problems,  examples,  and 
exercises.  The  problems  are  of  a  general  character  and 
are  accompanied  by  a  full  solution.  The  examples  are 
numerical,  and  are  intended  to  be  special  applications  of 
the  formulas  and  principles  contained  in  the  chapter  of 
which  they  form  a  part.  The  exercises  are  a  novel  feature 
of  the  work.  They  are  intended  particularly  to  draw  out 
and  fix  in  the  mind  the  general  principles  of  the  subject. 
The  answers  to  the  questions  under  this  head  are  not 
always  explicitly  given  in  the  text,  but  the  principle  in- 
volved in  the  answer  is  sufficiently  explained  there.  Ad- 
ditional questions  will  doubtless  suggest  themselves  both 
to  the  teacher  and  student,  and  in  some  cases  conditions 
may  be  added  to  those  given  in  the  exercise.  Thus,  in 
the  5th  Exercise,  on  page  23,  the  question  may  be  raised 
whether  the  weight  of  the  rope  is  to  be  considered ;  and 
if  so,  whether  the  velocity  is  to  be  uniform  or  variable ; 
also,  in  the  latter  case,  whether  the  acceleration  be  increas- 
ing or  decreasing. 

The  abstract  relations  which  exist  between  a  force  and 
the  motion  which  it  produces  in  a  body,  are  considered 
early  in  the  work.  I  do  not  consider  this  order  as  in 
the  least  essential,  but  I  have  usually  presented  the  sub- 
ject in  this  way  to  my  classes,  regardless  of  the  order 
given  in  the  text-book.  No  part  of  abstract  mechanics 
possesses  a  greater  interest  than  this,  and  the  questions 
pertaining  to  momentum  and  energy  which  grow  out  of 
these  relations  have  provoked  a  great  deal  of  discussion 
among  students  in  mechanics.  In  regard  to  Momentum 


PREFACE.  V 

and  Vis  Viva,  much  of  the  difficulty  which  arises  in  the 
mind  of  the  student  in  regard  to  them  would  be  removed, 
if  they  are  considered,  as  they  really  are,  mutually  inde- 
pendent of  each  other,  having  no  common  unit  between, 
them,  but  each  having  its  own  peculiar  unit. 

The  demonstration  of  the  formula  for  centrifugal  force 
may  appear  to  be  unnecessarily  lengthy,  but  I  trust  that 
the  student  will  gain  a  clearer  conception  of  the  mode  of 
action  of  the  forces,  and  be  better  satisfied  with  the  logic 
of  the  demonstration,  by  following  the  proof  here  given, 
than  by  certain  of  the  shorter  methods.  The  demonstra- 
tion by  some  of  the  latter  methods  is  defective,  although 
the  same  result  is  reached  by  them. 

The  principles  of  energy,  which  plays  such  an  important 
part  in  modern  physics,  have  been  explained,  and  the  prin- 
ciples of  both  Kinetic  and  Potential  Energy  used  in  thp 
solution  of  problems. 

HOISOKEN,  December,  1877. 


CONTENTS. 


The  numbers  of  the  Articles  and  the  leading  topic  are  placed  at  the  head  of  the  page.] 
CHAPTER   I. 

KINEMATICS. 

Motion. — Rest. — Kinematics. — Rest  and  motion. — Path. — Velocity.-— 
The  unit  of  velocity. — Variable  velocity. — Geometrical  illustration, 
— General  formula. — Circular  path. — Angular  velocity. — Resolved 
velocities. — Parallelogram  of  velocities. — Triangle  of  velocities. — 
Polygon  of  velocities. — Parallelopipedon  of  velocities. — Harmonic 
motion. — Periodic  motion. — Rotary  motion.  —  Problems. — Exer- 
cises.— Acceleration.  —Uniform  acceleration. —  Formulas. — Initial 
velocity. — Resultant  velocity. — Examples.  Articles  1-26. 

Pages  1-14. 

CHAPTER  II. 

KINETICS  (commonly  called  DYNAMICS). 

Matter. — A  body. — Force. — Names  for  Force. — Measure  of  foice. — 
Weight. — Standard  measures. — Force  represented  by  a  right  line. 
• — Poirit  of  application. — Inertia. — Molecular  motions. — Mechanics. 
— Kinetics. — Statics. — Molar  mechanics. — Molecular  mechanics. — 
Hydrostatics. — Pneumatics. — Thermodynamics. — Rotation. — Exer- 
cises.— Laws  of  Motion  : — FIRST  LAW — SECOND  LAW — THIRD  LAW. 
—  Parallelogram  of  forces. — Resolved  forces. —  Components  of 
force. — Resultant. — Value  of  the  resultant. — Rectangular  compo- 
nents. —  Resultant  of  conspiring  forces.  —  Exercises.  —  Constant 
force. — Constant  acceleration. — Constant  moving  force. — Illustra- 
tion,—  Normal  action. —  Force  of  gravity. — Law  of  UNIVERSAL 
GRAVITATION. — Gravity  constant  at  any  place. — Find  the  accelera- 
tion due  to  gravity.— Gravity  varies  with  the  latitude.— Variable 
weight,  causes  of.  —  Atmospheric  resistance.  —  Formulas  for  fail- 


Vlll  CONTENTS. 

ing  bodies  ;  Initial  velocity  up  or  down.  — r  Problems  :  Spherical 
shell ;  Hollow  sphere ;  Gravity  varies  as  the  distance  from  the  cen- 
tre of  the  earth;  Weight. — Mass,  measure  of. — Unit  of  mass. — 
Measure  of  weight. — Density. — ABSOLUTE  MEASURE  OF  FORCE.— 
Effective  force. — Formulas  for  velocity  and  space  when  the  accel- 
eration is  constant. — Problems. — Examples.  Articles  27-90. 

Pages  15-51. 

CHAPTER  III. 

WORK.      FRICTION. 

Definition  of  work ;  measure  of  ;  variable. — Motion  not  work. — Unit  of 
work.  —  Geometrical  illustration.  —  Work  independent  of  time  ; 
Time  and  velocity  implied. — Dynamic  effect. — Unit  of  dynamic 
effect. — Useful  and  prejudicial  work. — Work  when  the  force  acts 
obliquely  to  the  path. — Work  in  a  moving  body  (or  energy). — Fric- 
tion ;  experiments  on ;  angle  of ;  laws  of ;  coefficient  of  .—Prob- 
lems.— Examples. — Exercises.  Articles  91-110 Pages  52-65. 

CHAPTER  IV. 

ENERGY. 

Definition;  kinetic;  potential. — Heat  and  work. — Dynamical  theory  of 
heat. — MECHANICAL  EQUIVALENT  OF  HEAT.  Joule's  experiment. 
—  Transmutation  of  energy. —  Conservation  of  energy. —  Equili- 
brium of  energies. — Perpetual  motion. — Examples.  Articles  111- 
121 Pages  66-77. 

CHAPTER  V. 

MOMENTUM. 

Definition. — Impulse. — Not  a  force  .—Instantaneous  force. — Problems. 
— Impact. — Elasticity;  coefficient  of. — Elongation  of  a  prismatic 
bar. — Modulus  of  restitution. — Impact ;  non-elastic  bodies ;  loss  of 
velocity ;  of  perfectly  elastic  bodies ;  Discussion :  imperfectly 
elastic  bodies;  loss  of  kinetic  energy. — Examples. — Exercises.— 
Work,  momentum,  and  energy  compared.  Articles  122-189. 

Pages  78-93 


CONTENTS.  IX 

CHAPTER  VI. 

COMPOSITION  AND  EESOLUTION  OF  PRESSURES. 

Remark. — Resultant  pressure;  component. —  Two  equal  pressures; 
three  equal  pressures  having  equal  angles  -with  each  other. — Parallel- 
ogram of  pressures. — Direction  of  resultant;  value  of. — Triangle 
of  pressures  ;  converse  ;  proportional  to  sines.  — One  force  and  two 
directions  given. — Case  of  three  forces  in  one  plane  on  a  rigid  body. 
— Polygon  of  pressures. —  Examples. — Exercises. —  Resolution  of 
forces. — Rectangular  components  ;  angles  «  and  /3  ;  any  number  of 
forces;  resultant  of. — Direction  of  resultant. — Examples. — Forces 
referred  to  three  axes.  Articles  140-161 Pages  94-103. 

CHAPTER  VIL 

MOMENTS   OF    FORCES. 

Axis  of  rotation. — Moment  of  a  force  ;  force  oblique. — Axis  of  moments. 
— Definitions. — MOMENT  represented  by  triangle  ;  sign  of  ;  repre- 
sented by  its  axis  ;  composition  of. — Moments  of  three  concurrent 
forces  in  equilibrium. — Unit  of  moments. — Origin  of  moments. — 
Arm  in  terms  of  two  coordinate  axes. — Origin  of  moments. — Par- 
allel forces. — Problems. — Choice  of  origin. — Problems. — Couples; 
produce  rotation ;  moment  of  ;  equilibrium  of ;  resultant  of.  — Two 
contrary  couples  on  a  rigid  body. — A  force  resolved  into  a  couple 
and  another  force. — If  the  origin  is  on  the  resultant  of  any  number 
of  forces,  the  sum  of  their  moments  will  be  zero. — THREE  PARAL- 
LEL FORCES. — A  force  and  couple. — Remark. — Forces  in  one  plane 
if  not  in  equilibrium  are  equivalent  to  a  single  couple,  or  to  a  cou- 
ple and  force. — If  the  sine  of  the  moments  in  reference  to  three 
points  not  in  a  right  line  are  zero,  the  forces  will  be  in  equilibrium. 
— Problems. — Examples. — Exercises.  Articles  163-196. 

Pages  104-126. 

CHAPTER  VIII. 
PARALLEL  FORCES. 

Definition, — Resultant. — Centre  of  parallel  forces  ;  to  find  centre  ;  re- 
ferred to  three  axes. — Centre  of  mass. —  Examples. — Exercises 
Articles  197-203 Pages  127-134 


X  CONTENTS. 

CHAPTER  IX. 

CENTRE   OP  GRAVITY   OF  BODIES. 

Gravity  considered  as  parallel  forces  ;  centre  of  gravity  same  as  centre  of 
parallel  forces. — Resultant  equals  the  weight. — Body  supported; 
same  vertical. — Stable  equilibrium;  unstable;  indifferent. — Trial 
methods. — Examples. — Exercises. — Centre  of  gravity  of  a  part  of  a 
body  ;  of  several  bodies. — Use  of  coordinate  axes. — Examples. — 
Straight  line. — Perimeter  of  a  triangle. — Symmetrical  figures. -- 
Circular  arc. — Plane  triangle. —  Symmetrical  areas.  —  Irregular 
figures. — Zone. — Examples. — Triangular  pyramid. — Any  pyramid 
or  cone. — Spherical  sector. — Segment  of  a  sphere. — Examples. — 
THEOREMS  OF  PAPOS. — Centre  of  gravity  of  a  circular  arc. — Volume 
computed. — Examples.  Articles  204-335 Pages  135-154. 

CHAPTER   X. 
ENERGY. 

Energy  in  three  states. — POTENTIAL  ENERGY. — Equilibrium  of  a  rod 
resting  against  a  vertical  plane  and  a  curve  ;  against  a  pin  and 
curve. — Centre  of  gravity  lowest  in  catenary.— Maximum  surface 
generated  by  the  revolution  of  a  line  of  given  length. — Two  cylinders 
resting  within  another. — Equilibrium  of  a  body  on  an  inclined 
plane;  pulley;  straight  lever;  bent  lever. — KINETIC  ENERGY  of 
a  falling  body  ;  of  bodies  on  two  inclined  planes  ;  of  one  body  on  an 
inclined  plane  and  the  other  in  a  vertical  plane. — Movement  of 
bodies  on  a  pulley. — Discharge  of  a  fluid  through  an  orifice. — 
Examples.  Articles  236-250 Pages  155-169. 

CHAPTER   XI. 
CONSTRAINED  EQUILIBRIUM. 

Definition. — Normal  resultant. — Equilibrium  on  a  smooth  inclined 
plane;  on  rough  inclined  plane. — An  eccentric  disc. — Equilibrium 
of  a  beam  on  two  smooth  inclined  planes. — Equilibrium  of  a  beam 
in  a  smooth  bowl.  — Equilibrium  of  a  particle  on  the  arc  of  a  circle  ; 
on  the  arc  of  a  parabola  ;  on  arc  of  an  ellipse  ;  on  arc  of  a  hyper- 
bola.— Examples. — Exercises.  Articles  251-263 Pages  170-181 


CONTENTS.  XI 

CHAPTER  XII. 
ANALYTICAL  METHODS. 

Definitions. — Forces  not  in  one  plane;  general  case;  resolved.  - -Mo 
ments  of  resolved  forces.— Problems. — Tension  of  a  loaded  chord. — 
Weight  supported  by  a  string  and  strut. — Solution  of  a  problem  of 
rafters  analytically;  synthetically. — Examples. — Exercises.     Arti- 
cles 264-273 Pages  182-193. 

CHAPTER  XIII. 

STRENGTH  OP   BARS  AND   BEAMS. 

Strength  of  a  prismatic  piece. --Modulus  of  tenacity. — Formulas  for 
strength. — Law  of  resistance  of  beams. — Modulus  of  rupture. — 
Formula  for  moment  of  resistance. — Problems. — Examples. — Arti- 
cles 274-282 Pages  194-201. 

CHAPTER   XIV. 
MOTION  OF  A  PARTICLE  ON  AN   INCLINED  PLANE. 

Formulas. — Initial  velocity. — Problems:  motion  down  the  chord  of  a 
circle  ;  straight  line  of  quickest  descent  from  a  point  to  a  circle  ; 
from  one  circle  to  another ;  from  a  circle  to  a  line  ;  from  a  line  to 
a  circle. — Time  down  a  rough  plane  ;  approximate  formulas  ;  equa- 
tions adapted  to  the  movement  of  cars  on  an  inclined  plane. — 
Examples.  Articles  283-292 Pages  202-210. 

CHAPTER  XV. 

PROJECTILES. 

Definitions  ;  path  and  range.  — Path,  a  parabola ;  velocity  at  any  point 
of  ;  position  of  focus;  referred  to  rectangular  axes.  —The  range. — 
Time  of  flight. — Greatest  height. — Greatest  range. — Angle  of  eleva- 
tion for  greatest  height. — Equal  ranges. — Range  on  an  inclined 
plane. — Problems. — Examples. — Exercises.  Articles  293-30(5. 

Pages  211-219 


Xll  CONTENTS. 

CHAPTER  XVL 

CENTRAL   FORCES. 

Definition. — Centripetal  force;  nature  of. — The  orbit;  to  find. — Cen 
trifugal  force. — Value  of  centrifugal  force ;  in  terms  of  angttla* 
velocity ;  periodic  time. — Centrifugal  force  of  extended  masses. — 
Centrifugal  force  at  the  equator ;  when  equal  to  the  weight ;  value 
at  the  moon ;  on  railroad  curves. — Elevation  of  the  outer  rail. — 
The  conical  pendulum. — Examples. — Exercises.  Articles  307-323. 

Pages  220-235. 

CHAPTER    XVII. 

FORCES  WHICH  VARY  DIRECTLY  AS  THE  DISTANCE  FROM  THE 
CENTRE  OF  THE  FORCE. 

Remark. — Velocity  at  any  point  in  the  path. — Time  of  movement. — 
Problems  :  Simple  pendulum  ;  Compound  pendulum  ;  Length  of 
the  second's  pendulum ;  Time  lost  by  clock  carried  to  a  height ; 
Movement  of  a  body  through  the  earth ;  Vibrations  of  an  elastic 
bar.  —Examples.  Articles  324-332 Pages  236-249. 

CHAPTER  XVIII. 

GENERAL  PROPERTIES  OF  FLUIDS. 

Definition  of  fluids ;  liquids ;  uniform  bodies. — Forces  in  the  three 
states  of  matter. — Law  of  equal  pressures. — Normal  pressures. — 
Equal  transmission  of  pressures. — Vertical  pressures. — Pressure  on 
the  base  of  a  vessel. — Pressure  on  the  inside  of  a  vessel. — Resolved 
pressures  in  any  direction ;  on  an  immersed  body. — Point  of  appli- 
cation of  resultant. — Equilibrium  of  fluids  of  different  densities.— 
Examples.— Exercises.  Articles  333-348 Pages  250-259. 

CHAPTER  XIX. 

SPECIFIC  GRAVITY. 

Definition. — Specific  gravity  of  a  body  more  dense  than  water ;  less 
dense  than  water ;  of  a  liquid.  — Absolute  weight  of  a  body. — Speci- 
fic gravity  of  a  soluble  body  ;  of  air. — Hydrometers  — Definition. — 


CONTENTS.  Xlll 

Areometer  of  constant  weight. — Nicholson's  Hydrometer. — Prob- 
lems ;  mechanical  combinations ;  chemical  combinations ;  specific 
gravity  of  a  body,  the  weight  in  air  being  given. — Examples. — 
Exercises.  Articles  349-362 Pages  260-271. 

CHAPTER   XX. 

HYDROSTATICS. 

Compressibility  of  liquids. — Free  surface. — Level  surface. — Problems. 
— Examples. — Law  of  pressure. — Pressure  against  a  rectangle,  the 
upper  end  being  in  the  free  surface ;  being  entirely  submerged ; 
against  any  surface. — Problems. — Examples. — CENTRE  OP  PKKS- 
8URE  ;  of  a  rectangle  ;  submerged  rectangle  ;  triangle,  base  down  ; 
base  up. — Plane  of  flotation. — Condition  of  equilibrium  of  a  body  ; 
stable  equilibrium. — Depth  of  flotation. — Problems. — Examples. — • 
Articles  363-379 Pages  272-288. 

CHAPTER  XXI. 

HYDRODYNAMICa 

Mean  velocity. — Permanent  flow. — Variable  velocity. — Velocity  of  dia 
charge  from  an  orifice  in  the  base  of  a  vessel ;  from  an  orifice  in 
the  side. — Head  due  to  the  velocity. — Vertical  pressure  on  the  free 
surface. — Pressure  of  the  air. — A  vertical  jet. — Several  orifices  in 
the  side  of  a  vessel. — Oblique  jet. — Coefficients  of  contraction  ;  of 
velocity ;  of  discharge. — Discharge  through  a  large  orifice ;  external 
pressure"  considered.  —  A  weir. — Rectangular  notch,  quantity  of 
flow. — Mean  velocity  through  a  weir. — Coefficient  of  discharge 
through  a  weir. — Submerged  rectangular  orifice. — Flow  through 
long  pipes. — Formulas  for  circular  pipes  ;  diameter  of  pipe. — Effect 
of  the  condition  of  the  pipe.  —Flow  in  rivers  and  canals  ;  character 
of  the  bed  of  the  stream.  — Form  of  free  surface  in  the  cross  section 
of  a  stream. — Backwater. — Backwater  in  rapid  streams.  — Problems. 
—Examples.— Exercises.  Articles  380-409 Pages  289-311. 

CHAPTER   VTTT. 

GASES   AND    VAPORS. 

Definition. — Pressure  of  the  atmosphere,  Torricellian  experiment. — 
Barometer. — Height  of  a  homogeneous  atmosphere. — Boyle's  (or 


XIV  CONTENTS. 

Mariotte's)  law  ;  law  not  perfect. — Manometers. — Expansion  of  gas 
due  to  temperature. — Coefficient  of  expansion  of  air  ;  of  perfect 
gas. — Volume  of  a  gas  for  given  temperature  and  pressure. — Ther- 
mometers.—  Compressing  air. —  Steam  or  vapors. —  Problem*. — 
Weight  of  a  cubic  foot  of  air  for  any  temperature  and  pressure  ; 
Weight  of  steam;  Spherical  air- bubble  rising  through  water. — 
Exhaustion  by  an  air-pump. — To  measure  elevations  with  a  baro- 
meter.— Flow  of  a  gas  into  a  vacuum. — Examples. — Examples  from 
the  examination  papers  for  the  University  of  London.  Articles 
410-434 Pages  312-335 

APPENDIX   i 

TABLE      L — Experiments  on  friction  without  unguents 337-340 

"       II. — Experiments  on  the  friction  of  unctuous  surfaces.  341 
*'      III. — Experiments   on   friction   with   unguents  inter- 
posed    342-344 

44      IV.— Specific  gravity  of  bodie* 344-347 


MEASURES.  XV 

FRENCH  AND  ENGLISH  MEASURES. 

A  DECIMETRE  DIVIDED  INTO  CENTIMETRES  AND   MILLIMETRES. 


* 


Inches  and  tenths. 

FRENCH  MEASURES  IN  EQUIVALENT  ENGLISH  MEASURES. 

MEASURES  OF   LENGTH. 

1  Millimetre   =  0' 03937079  inch,  or  about  ^  inch. 
1  Centimetre  =  0" 3937079  inch,  or  about  0'4  inch. 
1  Decimetre   =  3  "937079  inches. 
1  Metre  —  39 '37079  inches  =  3  "28  feet  nearly. 

1  Kilometre    =  39370 '  79  inches. 


MEASURES  OP  AKEA. 
1  Bq.  millimetre    =  0' 00155006  sq.  inch. 
1  sq.  centimetre  =  0  155006  sq.  inch. 
1  sq.  decimetre    =  15  •  5006  sq.  inches. 
1  sq.  metre  =  1550 '06  sq.  inches,  or  10.764  sq.  fe«l 


MEASURES  OP  VOLUME. 

1  cu.  centimetre  =  0' 610271  cu.  inch. 

1  cu.  decimetre    =  61 -  0271  cu.  inches. 

1  cu.  metre          =  61027 '1  cu.  inches. 

The  litre  (used  for  liquids)  is  the  same  as  the  cubic  decimetre. 


MEASURES  OP  WEIGHT. 

1  Milligramme    =  0"  015432349  grain. 
1  Centigramme  =  0' 15432349  grain. 
1  Decigramme    =  1  •  5432349  grains. 
1  Gramme  =  15 '432349  grains. 

1  Kilogramme    =  15432 '349  grains,  or  2.2  Ibs.  nearly. 


TWO   UNITS  INVOLVED. 

1  Gramme  per  sq.  centimetre          =  2' 048098  Ibs.  per  sq.  foot. 
1  Kilogramme  per  sq.  metre  =  0' 2048098     "  " 

1  Kilogramme  per  sq.  millimetre    =  2-048098       "  " 

1  Kilogramme  metre  =  7 '23314  foot-pounds. 

=  7^  foot-pounds  nearly. 

1  force  de  cheral  =  75  kilogrammetres  per  second,  or  542£  f oot-pot\jida 
per  second  nearly.     1  horse-power  =  550  foot-pounds  per  second. 


JCVl 


MEASURES. 


ENGLISH  MEASURES  IN  EQUIVALENT  FRENCH  MEASURES 


MEASURES  OF  LENGTH. 
1  inch  =  25  •  39954  millimetres. 
1  foot  =  0-304794  metre. 
1  yard  =  0' 9143834  metre. 
1  mile  =  1-60932  kilometre. 


MEASURES  OF  AREA. 
1  sq.  inch  =  645 '137  sq.  milli'tres. 
1  sq.  foot  =  0-0929  sq.  metre. 
1  sq.  yard  =  0"  83609  sq.  metre. 
1  sq.  mile  =  2 '59  sq.  kilometres. 


MEASURES  OF   CAPACITY. 
1  pint      -  0-5676  litre. 
1  gallon   =  4 -5410  litres. 
1  bushel  =  36-3281  litres. 


MEASURES  OF  VOLUME, 
1  cn.  inch  =  16386  6  ou.  milli'tres. 
1  cu.  foot  =  0-0283  cu.  metre. 
1  cu.  yard  =  0  7645  cu.  metre. 


MEASURES  OF  WEIGHT. 

1  grain         =  0-064799  gramme. 
1  oz.  avoir.  =  28 "  3496  grammes. 
1  Ib.  avoir.  =  0'4535  kilogramme. 
1  ton  =  1-01605  tons. 

-  1016-05kilog. 


TWO   UNITS  INVOLVED. 


1  Ib.  per  sq.  foot  =  4 '88261  kilog.  per  sq.  metre. 
1  Ib.  per  sq.  inch  =  0'0703  kilog.  per  sq.  centimetre. 
1  foot-pound         =  0*1382  kilogrammetre. 


Letters. 
A  a 

Names. 

Alpha 

B/3 

Beta 

r  v 

Gamma 

A  8 

Delta 

£  f 

Epsilon 

z  C 

Zeta 

Hf 

Eta 

-ead 

Theta 

GREEK 

ALPHABET. 

Letters. 
I    i 

Names. 
I5ta 

K     K 

Kappa 

A  X 

Lambda 

M  n 

Mu 

N  v 

Nu 

3£ 

Xi 

0  o 

Omicron 

n  IT 

Pi 

Letters. 
P   p 

Names. 
Rho 

2  o-  s 

Sigma 

r  T 

Tau 

Y  v 

Upsilon 

*  <i> 

Phi 

x* 

Chi 

*    1/r 

Psi 

O    Cl) 

Omega 

ELEMENTARY   MECHANICS, 


CHAPTER  I. 

KINEMA.TICS   OR   MOTION. 

•*"*  Xta       ti-t 

Definitions. 

1.  Motion. — As  we  look  about  us  we  see  many  objects 
in  motion ;  such  as  men  walking,  birds  flying,  ships  sail- 
ing, etc.     We  also  know  that  the  earth  and  other  planets 
are  moving  through  space. 

2.  Rest. — We  also  see  many  objects  apparently  at  rest ; 
that   is,   they  remain  in  the  same  place  in  reference  to 
surrounding  objects.     Thus,  hills,  rocks,  buildings,  etc., 
appear  to  be  at  rest. 

3.  Kinematics  is  the  science  of  motion. — It  does  not 
consider   the   cause  of  the   motion,   but   determines    its 
measure,  and  the  relations  between  different  motions. 

4.  Rest  and  motion  are  relative  terms. — A  body  may 
be  at  rest  in  reference  to  some  objects,  and  in  motion  in 
reference  to  others.     Thus,  a  person  sitting  on  the  deck 
of  a  ship  maj  be  at  rest  in  reference  to  the  objects  on 
the  ship,  while  he  is  moving  with  the  ship  over  the  water. 
But  if  he  should  run  toward  the  stern  at  the  same  rate 
that  the  vessel  is  advancing,  he  will  appear  to  be  at  rest 
in  reference  to  objects  on  the  shore,  and  moving  in  refer- 
ence to  the  objects  on  deck.     Objects  at  rest  on  the  earth 
are  *noving  through  space  with  the  earth  at  the  rate  of 
more  than  67,000  miles  per  hour. 


2  KINEMATICS.  [5,6.] 

The  motion  of  one  body  in  reference  tc  another  also  in 
motion  is  called  relative;  but  in  reference  to  a  fixed 
object  it  is  called  absolute  or  actual. 

,'5.  The  path  of  a  body  is  the  line  traced  by  its  central 
point. — If  all  the  points  of  a  body  move  in  parallel  lines, 
any  one  of  the  lines  may  be  taken  as  the  path.  Unless 
otherwise  stated,  we  will  assume  that  the  body  is  reduced 
to  a  mere  particle.  The  path  is  also  called  the  space  over 
which  a  body  moves. 

Velocity. 

6.  Velocity  is  rate  of  motion. — When  a  body  passes 
over  equal  successive  portions  of  space  in  equal  times,  its 
rate  is  uniform.  In  all  other  cases  it  13  variable.  Thus, 

if   a   body    moves    uniformly 

1334? 

•A' ' ' •— >s     from  A  to  B  in  four  seconds, 

the   spaces   passed   over  each 

second  will  be  one-fourth  of  AB,  or  A-l  —  1-2  =  2-3,  etc., 
and  any  one  of  these  spaces  is  the  velocity. 

When  the  motion  is  uniform,  the  velocity  is  the  space 
passed  over  in  a  unit  of  time,  and  the  velocity  is  said  to 
be  constant. 

If  s  =  the  space  passed  over  uniformly, 
t  =•  the  time  of  the  movement,  and 
v  =  the  velocity, 
then  we  have,  according  to  the  definition, 

v  =  -t-,       .        .  (1) 

from  which  we  find, 

.        .  (2) 

and 


[»-».]  VELOCITY.  3 

7.  The  unit  of  velocity  is  understxtod  to  be  one  fooi 
per  second,  unless  otherwise  stated.      If  other  units  are 
given,  their  equivalent  value  may  be  found  in  feet  per 
secoTid. 

8.  Variable  Velocity  is  that  in  which   the  rate  of 
motion  is  constantly  changing. — The  true  measure  in  this 
case  cannot  be  the  space  passed  over  during  any  single 
second,  but  it  is  the  space  which  would  be  passed  over 
during  a  second  if  the  body  moved  uniformly  at  the  rate 
which  it  had  at  the  instant  considered. 

We  are  familiar  with  this  fact.  We  say  that  a  train  of 
care  moved  at  the  rate  of,  say,  forty  miles  per  hour,  when 
it  may  have  moved  at  that  rate  for  an  instant  only,  and  in 
coming  to  rest  it  may  have  moved  at  all  conceivable  rates 
less  than  forty  miles  per  hour. 

9.  Geometrical  Illustration. — Variable  velocity  may 
be  represented   by  a  figure.     Thus,  in  Fig.  2,  let  AB 
represent  the   time,   say  four  seconds. 

Divide   it  into   four  equal  parts,  each 

of  which   will   represent    one    second. 

At  the  several  points  of  division  erect 

ordinates,  and  make  them  proportional 

to  the  velocities  corresponding  to  those 

times  ;  the  ordinate  a\  representing  the 

velocity  at  the  end  of  one  second  ;  b2  at 

the  end  of  two  seconds,  and  so  on.    A  FlG'2' 

curve  AabcOm&j  be  drawn  through  the  extremities  of  these 

ordinates,  such  that  the  ordinate  at  any  point  will  represent 

the  velocity  corresponding  to  that  time.      If  a  line  ad  be 

drawn  through  a  parallel  to  A£,  the  number  of  square 

units  in  ~Lad%  will  also  represent  the  velocity ;  for  the  side 

1-2  of  the  rectangle  being  unity,  there  will  be  the  same 

number  of  square  units  in  the  rectangle  that  there  are 


4:  KINEMATICS.  [10,  1JJ.J 

linear  units  in  al,  and  similarly  for  any  other  part  of  the 
figure. 

The  area  of  the  figure  ABC  \fi\\  represent  the  number 
of  units  of  space  passed  over  by  the  body  in  four 
seconds. 

10.  The  general  computations  for  variable  velocity 
belong  to  higher  mathematics  ;  but  we  are  enabled  to 
treat  of  some  cases  in  an  elementary  manner,  as  will  be 
shown  hereafter. 

The  velocity  may  be  found  in  practical  cases  with  suffi- 
cient accuracy  by  finding  the  actual  space  passed  over  by 
a  body  in  a  very  short  space  of  time,  and  considering  the 
motion  as  uniform  during  that  time. 

If  At  (read  element  of  time,  or  simply  delta  I)  be  the 
element  of  time,  and 

ds  =•  the  corresponding  space, 
then,  according  to  the  definition,  we  have 


v  = 


At 


11.  The  path  may  be  the  arc  of  a  circle,  or  any  other 
curved  line,  in  which  case  the  space  will  be  the  length  of 
the  developed  line.     If  AB  be  the 
arc  of  a  circle  passed  over  in  time  t, 
we  have  v  =  AB  -~  t,  as  before. 

12.  Angular  Velocity  is  the  rate 
of  angular  movement.  It  is  measur<  <l 
by  the  circular  arc  having  unity  for  its 
radius,  which  would  be  generated  by 
the  extremity  of  the  radius  if  it  turned 
about  the  centre  at  the  same  rate  as  that  of  the  body.     If 
AB  =  v,  and  Oa  =  1,  then  ab  —  angular  velocity. 


[13.]  VELOCITY.  5 

Or,  if  AB  =  8  =  the  space  passed  over  uniformly  by 

the  body, 
t  =  the  time, 

r  =  OA  =  the  radius  of  the  arc  AB,  and 
w  =  the  angular  velocity, 
then 

o 

—  =  the  velocity  along  AS,  and 
t 


t.r 

13.  Resolved  Velocities.  —  Suppose  that  a  body  moves 
uniformly  from  A  to  B.     At  the  extremities  of  the  line 
draw  two  lines,  A  G  and   BC, 
forming    a    right    angle    at    C. 
Then,  if  a  body  moves  from  A 
to  6yin  the  same  time  that  one 
moves  from  A  to  B,  the  velocity 
of  the  former  will  equal  the  latter  into  cos.  BAO. 

Or,  if 

v  =  the  velocity  along  AB, 
Vi  =  the  velocity  along  A  C,  and 

£  =  angle  B  >AG\ 
then, 

Vi  =  v  cos.  /9.    .        .        .        (1) 
Similarly, 

<vz  =  v  sin.  /S.    .        .        .        (2) 

when  v2  is  the  uniform  velocity  along  BC. 

Suppose  that  a  body  is  pushed  in  a  due  southerly  direc- 
tion, parallel  to  BC,  and  at  the  same  time  in  a  westerly 
direction  parallel  to  OA  ;  if  the  velocities  in  these  direc- 
tions are  uniform  and  proportional  to  BC  and  CA  re« 
spectively,  the  resultant  motion  will  be  along  a  line  paral= 
lei  to  BA,  and  the  velocity  will  be  proportional  to  BA. 


6  KINEMATICS.  [14,15.j 

The  velocities  represented  by  BC  and  OA  are  called 
component  velocities,  and  that  by  BA  is  called  the  result- 
ant velocity. 

14.  Parallelogram  of  Velocities.  —  The  component 
velocities  AC  and  BC  may  make  any  angle  with  each 
other.  Thus,  if  in  Fig.  5 

v  =  the  velocity  along  AB, 
vt  =  the  velocity  along  A  C, 
vz  =  the  velocity  along  AD, 
0  =  the  angle  D  AC; 

then,  from  plane  trigonometry,  we  have 


AB  =  VAC*  +  AI?  +  1AC.AB  cos.  DAG-, 

,  _ 

v  =  tfj£+  vj  +  2y^2  cos.  £. 

If  the  angle  DACi&  obtuse,  cos.  /9  will  be  negative.   If 

DA  C  is  a  right  angle,  we  have 
cos.  90°  =  0,  and 


This  principle  may  be  stated 
as  follows : 

~ 


If  two/^velocities  are  repre- 
sented in  magnitude  and  direction  by  the  adjacent  sides 
of  a  parallelogram,  the  resultant  velocity  will  be  repre- 
sented in  magnitude  and  direction  by  that  diagonal  of  the 
parallelogram  which  lies  between  these  sides. 

15.  Triangle  of  Velocities. — If  two  concurrent  veloci- 
ties be  represented  in  magnitude  and  direction  by  two  sides 
of  a  triangle  taken  in  their  order,  the  resultant  velocity  will 
be  represented  in  magnitude  and  direction  by  the  third 
side. 


[16-18.] 


VELOCITY. 


Thus,  in  Fig.  X5,  if  AC  represent  the  velocity  »,.  in  the 
direction  from  A  towards  C,  and  CB  represent  the  velo- 
city Vfr  acting  from  G  towards  B,  then  will  AB  represent 
the  resultant  velocity.  This  is  evident  from  the  preced- 
ing article. 

16.  Polygon  of  Velocities.— If  several  velocities^  act- 
ing successively,  carry  a  body  around  a  polygon,  they  will 
produce  rest  if  they  all  act  at 

the  same  time.  And,  if  several 
velocities  be  represented  in  mag- 
nitude and  direction  by  the 
successive  sides  of  a,  polygon 
taken  in  their  order,  when  they 
act  all  at  the  same  time  their 
resultant  velocity  will  be  repre- 
sented by  the  closing  side  of  the  polygon. 

This  is  proved  by  means  of  the  preceding  article,  by 
compounding  two  velocities,  then  their  resultant  and  a 
third  velocity,  and  so  on  to  the  last. 

17.  Parallelopipedon  of  Velocities. — If  three  veloci- 
ties not  in  one  plane  be  represented  by  the  three  adjacent 
edges  of  a parallelopipedon,  the  result- 
ant velocity  will  be  represented  by  that 

diagonal  which  passes    through    the 
common  point  of  the  three  edges. 

This  may  also  be  proved  by  the 
Triangle  of  Velocities.  The  resultant 
velocity  of  a  body  which  has  a  south- 
ward, westward,  and  downward  mo- 
tion, will  be  southwesterly  and  downward. 

18.  Harmonic  Motion. — If  a  body  moves  at  a  uniform 
rate  around  the  circumference  of  a  circle,  the  foot  of  the 
perpendicular  from   the  body  upon    the  diameter  will 


FIG.  7. 


8  KINEMATICS.  [10-81.1 

appear  to  move  to  and  fro  along  the  diameter  with  a  varia- 
ble velocity.     Thus,  in  Fig.  8,  if  the  point  move  uniformly 
around  the  circumference  A  CB,  the  foot  of  the  perpen- 
dicular 0  will  move  from  A  towards  B, 
thence  from  JS  towards  -4,  and  so  on. 
The  motion  of  the  point  O  is  said  to 
be  harmonic,  for  the  law  of  its  move- 
ment is  similar  to  that  of  a  musical 
string,  or  a  tuning  fork. 

19.  Periodic    Motion   is    that    in 
which  the  motion  repeats  itself. — Thus, 
in  Fig.  8,  the  point  that  moves  to  and  fro  along  the  diam- 
eter  AJB   has  a  periodic   motion.     A   pendulum,   as  it 
vibrates  to  and  fro,  is  another  example. 

20.  Rotary  Motion  is  motion  about  an  axis.     It  is 
measured  by  its  angular  velocity.     See  Article  12. 

The  point  about  which  it  moves  may  also  have  a  pro- 
gressive velocity.  Thus,  the  wheels  of  a  carriage  have  a 
rotary  motion  about  the  axles,  while  the  axles  have  a 
progressive  movement.  The  moon  revolves  about  the 
earth  as  a  centre,  and  the  earth  not  only  revolves  on  its 
axis,  but  also  revolves  around  the  sun. 

21.  Problems. — 1.  If  the  circumference  of  the  earth 
at  the  equator  is  25,000  miles,  what  is  the  velocity  in  feet 
per  second  of  a  point  on  the  equator? 

The  earth  turns  on  its  axis  once  in  twenty-four  hours, 
and  there  are  5,280  feet  in  a  mile ;  hence,  the  velocity  in 
feet  per  second  is 

25000  x  5280 

V  —  51 £n ZJA 

24:  x  60  x  60 
2.  Required  the  angular  velocity  of  the  earth. 


[21.]  VELOCITY.  9 

The  circumference  of  a  circle,  whose  radius  is  unity 
is  2-7T  =  6.28  +  ;  hence,  the  angular  velocity  per  hour  ia 


The  angular  velocity  per  hour,  in  terms  of  degrees  is 

1 
which  per  minute  is 

and  per  second  is 


360° 
w  =  -—-  =  15°; 

24 


15°      1° 
"=60   =5= 


EXERCISES. 

1.  If  one  body  moves  at  the  rate  of  10  miles  per  hour,  and  another 

body,  starting  from  the  same  place,  moves  in  an  opposite  direction 
at  the  rate  of  15  miles  per  hour,  both  moving  uniformly,  find  the 
distance  between  them  at  the  end  of  10  minutes. 

2.  Which  moves  faster :  a  body  moving  6  feet  per  second  or  one  moving 

100  yards  per  minute  ? 

8.  A  railway  train  travels  90  miles  in  two  hours  ;  find  the  velocity  in 
feet  per  second. 

4.  Two  bodies  start  from  the  same  place  at  the  same  time,  and  move 

uniformly  at  right  angles  with  one  another,  one  at  the  rate  of  8  feet 
per  second,  the  other  at  the  rate  of  15  feet  per  second  ;  find  the 
distance  between  them  at  the  end  of  one  second ;  also  at  the  end 
of  n  seconds. 

5.  If  a  train  moves  uniformly  at  the  rate  of  40  miles  per  hour,  how 

many  seconds  will  it  take  to  go  400  feet  ? 

6.  If  the  fly-wheel  of  an  engine  revolves  200  times  per  minute,  what 

will  be  its  angular  velocity  per  second  ? 

7.  If  a  person  aims  to  row  directly  across  a  stream  at  the  rate  of  3  miles 

per  hour,  while  the  stream  carries  him  downward  at  the  rate  of 
2  miles  per  hour,  at  what  rate  will  the  boat  move  ? 
1* 


10  KINEMATICS.  [22,23., 

8.  If  a  wheel  is  rolled  directly  across  the  deck  of  a  ship  at  the  rate  of 
15  feet  per  second,  while  the  ship  is  moving  10  miles  per  hour, 
find  the  velocity  of  the  wheel  in  space. 

0.  If  the  velocity  is  one  metre  per  hour,  find  the  velocity  in  feet  pei 
second. 

(One  metre  is  3.28  feet  nearly.) 

10.  Two  bodies  start  from  the  same  place  at  the  same  time,  and  mov« 
in  paths  which  are  inclined  60  degrees  to  each  other,  one  moving 
at  the  rate  of  5  feet  per  second  and  the  other  at  the  rate  of  10 
feet  per  second  ;  required  the  distance  between  them  at  the  end 
of  two  seconds. 


Acceleration. 

22.  Acceleration  is  the  rate  of  change  of  the  velocity. 
— If  the  rate  of  motion  be  uniform,  the  velocity  is  con- 
stant, and  there  will  be  no  acceleration.     If  the  velocity 
constantly  increases,  the  rate  of  change  is  called  a  positive 
acceleration,   but  if  it  constantly  decreases    it   is  called 
negative.     The  rate  of  change  may  be  uniform  or  con- 
stantly changing.     If  the  acceleration  is   uniform,  it  is 
measured  by  the  amount  by  which  the  velocity  is  increased 
in  a  unit  of  time  /  and  if  it  be  variable,  it  is  measured  by 
the  amount  by  which  the  velocity  would  be  increased  in  a 
unit  of  time  if  its  rate  of  increase  continued  the  same  as 
at  the  instant  considered.     The  unit  of  time   is  usually 
one  second.     If  the  velocity  is  decreasing,  the  same  defi- 
nition  applies  by   substituting  the   word  decreased  for 
increased. 

23.  Uniform  Acceleration  may  be  represented  by  a 
triangle.     Thus,  in  Fig.  9,  let  AB  represent  the  time — say 
four  seconds — and  BO  the  velocity  at  the  end  of  four 
seconds,  the  body  having  started  from  rest.     Divide  AB 
into  four  equal  parts  at  the  points  b,  g,  k,  and  draw  the 
horizontal  lines  be,  ge,  7ch,  to  the  line  drawn  from  A  to  C\ 


[24.J  ACCELERATION.  11 

then  will  bo  represent  the  velocity  at  the  end  of  the  first 
second,  ge,  at  the  end  of  the  second  second,  and  so  on. 
Draw  the  vertical  line  cd\  then  will  de  represent  the 
increase  of  the  velocity  during  the 
second  second,  and  hence  is  the  accel- 
eration. Similarly  fh  will  represent 
the  acceleration  during  the  third 
second,  and  iC  that  during  the  fourth. 
But  iO=fh  =  de  —  boj  or  the  velo- 
city increases  uniformly  and  the  accel- 
eration is  constant,  and  the  velocity  at 
the  end  of  the  first  second  equals  the 
acceleration. 

The  space  passed  over  during  the  second  second  will  be 
represented  by  the  trapezoid  bceg,  which  is  3  times  the 
triangle  Abe.  The  space  gehk  is  5  times  the  triangle  Abe. 
Hence,  generally,  if  the  times  are  represented  by  the 
natural  numbers 

1,        2,        3,        4,  etc.,  the  spaces  will  be 
1,        3,        5,        7,  etc. 

When  the  acceleration  is  uniform,  the  spaces  described 
may  be  laid  off  on  a  straight  line,  as  is  shown  by  the  line 
AC,  Fig.  11. 

24.  Formulas  for  uniform  acceleration. 
Let 

f=  the  uniform  acceleration, 
t  =  the  time, 

s  =  the  space  passed  over  in  the  time  t, 
v  =  the  velocity  at  the  end  of  the  time  t ; 

then,  in  Fig.  9,  we  have 


12 


or 


KINEMATICS. 

Ab  \lc\\AB\  BG\ 

i:/::<U; 

.'.  9  =ft.         .  , 

Area  ABC  =%BCxAB 


[25-J 


and  by  eliminating  t  we  have 


.          (1) 

.        (2) 
.        (3) 


Eliminating  v  between  equations  (1)  and  (2)  gives 

«  =  i/P.       ...        (4) 

When  any  two  of  the  quantities  /j  v,  t,  s  are  given,  the 
other  two  may  be  found  from  the  preceding  equations. 

25.  Initial  Velocity.  —  The  velocity  which  a  body  has 
at  the  instant  t  begins  to  be  reckoned,  is  called  initial 
velocity.  This  may  be  illustrated 
by  Fig.  10,  in  which  AB  repre- 
sents the  time,  DA  the  initial 
velocity,  EC  the  final  velocity, 
and  DECA  the  space.  It  thus 
appears  that  the  final  velocity  will 
be  that  due  to  the  acceleration 
plus  the  initial  velocity  ;  and  the 
space  will  be  that  due  to  a  uniform  movement  equal  to 
the  initial  velocity,  plus  that  due  to  the  acceleration  ; 
hence,  if 

V0  =  the  initial  velocity, 
then 

tv0  =  ABED  =  the  space  due  to  the  ini- 


Pio.  10. 


[26.] 


ACCELERATION. 


tial  velocity ;  and  using  the  same  notation  as  in  the  pre- 
ceding article,  we  have  for  this  case 

.        .  .        (1) 

.        .  .        (2)      ' 

.  -        (3) 


If  the  acceleration  is  decreasing,  we  have 

*  =  *>o  —ft,       ...        (4) 

•  =  fc>o-i/P.  -        -        (5) 

26.  The  Resultant  of  Variable  and  Constant  Velo- 
cities. —  In  Fig.  11,  let  the  horizontal  velocity  be  constant, 
and  the  vertical  motion  be  uniformly  accelerated.  If  Aa 
is  the  space  passed  over  in  a  horizon- 
tal direction  during  the  first  second, 
and  Ad,  the  vertical  space  during  the 
same  time,  then  will  the  body  be  at 
the  intersection  of  the  lines  drawn 
respectively  through  a  and  d\  the 
former,  vertical  and  the  latter,  hori- 
zontal. In  a  similar  way  its  position 
may  be  found  at  the  end  of  any  given 
time. 

The  locus  of  these  points  will  be 
the  path  of  the  body,  and  will  be  a  parabola. 

In  a  similar  way  the  path  may  be  found,  when  both 
velocities  are  variable  and  acting  at  any  angle  with  each 
other. 


FIG.  11. 


EXAMPLES. 


1.  If  a  body  starts  from  rest  and  has  a  uniform  accelera- 
tion of  5  feet  per  second,  how  far  will  it  move  in  10 
seconds,  and  what  velocity  will  it  acquire  ? 


14  KINEMATICS.  [26.  J 

2.  A  body   starts  from   rest  and  acquires  a  velocity   of 

100  feet  in  4  seconds ;  required  the  acceleration  and 
the  space  passed  over  during  the  first  second. 

3.  If  a  body  starts  from  rest  with  a  uniform  acceleration 

of  32  feet  per  second,  how  far  will  it  move  during 
the  4th  second  ? 

4.  A  body  has  an  initial  velocity  of  50  feet  per  second, 

and  has  a  velocity  of  100  feet  at  the  end  of  8  seconds  ; 
required  the  acceleration  in  feet  per  second. 

5.  If  the  acceleration  is. uniformly  32  feet  per  second; 

required  the  time  necessary  to  pass  over  200  feet, 
the  body  starting  from  rest. 

6.  If  the  acceleration  is  20  metres  in  two  minutes,  what 

will  be  the  acceleration  in  feet  per  second  ? 

7.  If  the  acceleration  is  32£  feet  per  second,  what  will  be 

the  value  in  metres  per  second  2 


CHAPTER  II. 

KINETICS  : 

(Commonly  called  Dynamics.) 

27.  Matter  is  the  substance  of  which  bodies  are  com- 
posed.— In  its  grosser  forms  we  gain  a  knowledge  of  it 
by  common  experience.     It  is  difficult,  if  not  impossible, 
to  define  it  so  that  a  person  who  is  not  already  familiar 
with  it  will  gain  a  correct  notion  of  it.     ]*t  has  certain 
properties,  such  as  extension,  divisibility,  porosity,  elas- 
ticity, etc.,  which  it  is  the  province  of  works  on  physics  to 
investigate.     For  the  purposes  of  Mechanical  Science,  it 
may  be  defined  as  that  upon  which  force  acts.     But,  when 
we  consider  the  effect  of  forces  upon  bodies,  it  is  necessary 
to  know,  or  to  assume,  certain  properties,  especially  the 
compressibility  and  elasticity  of  matter. 

28.  A  Body  is  a  finite  portion  of  matter. — An  atom  is 
an  indivisible  portion  of  matter.     It  is  an  ideal  tiling, 
since  we  know  nothing  of  its  essential  nature,  although  it 
has  been  a  subject  of  much  speculation.     It  is  assumed 
that  a  body  may  be  divided  and  subdivided,  until  parts 
might  be  reached  which,  from  their  constitution,  cannot 
be  again  divided.     A  molecule  is  the  smallest  portion  of  a 
body  which  can  exist  without  changing  its  nature.     It  is 
composed  of  two  or  more  atoms.     Thus,  a  molecule  of 
water  is  composed  of  two  atoms  of  hydrogen  and  one  atom 
of  oxygen,  and  if  they  be  separated  the  result  is  no  longer 
water  but  two  distinct  substances. 

A  particle  is  a  very  small  body,  or  a  small  portion  of  a 


J6  KINETICS.  [29,30.] 

body.  It  may  be  composed  of  several  molecules.  It  ha? 
no  reference  to  the  constitution  of  the  body.  In  mechan- 
ics it  is  considered  as  a  material  point. 

According  to  the  speculations  of  Sir  W.  Thomson, 
Maxwell,  Tait  and  others,  the  diameter  of  a  molecule  ex- 
ceeds ^250^000  of  an  inch,  and  is  less  than  ^^M  °^  an  inc^' 

In  order  to  give  an  idea  of  the  minuteness  of  a  mole- 
cule, Sir  W.  Thomson  states  that  if  a  body  of  the  size  of 
an  ordinary  pea  be  expanded  to  the  size  of  the  earth,  all 
the  molecules  expanding  in  the  same  ratio,  the  molecules 
would  be  between  the  sizes  of  fine  shot  and  a  cricket  ball. 


Force. 

29.  Force  is  that  which  changes  or  tends  to  charge  the 
state  of  a  body  in  reference  to  rest  or  motion. — ID  is  tv 
cause,  the  essential  nature  of  which  we  are  ignorant  of. 
We  deal  only  with  the  laws  of  ^ts  action.     These  laws  are 
determined  from  observation  combined  with  certain  com- 
putations.    All  forces  do  not  act  according  to  the  same 
laws.     Thus,  it  has  been  found  that  the  force  of  gravity 
varies  inversely  as  the  square  of  the  distance  from  the 
attracting  body ;  molecular  force  varies  directly  as   the 
distance  between  the  particles ;  and  other  forces  may  vary 
according  to  other  laws. 

30.  Forces  are  called  by  different  names,  according  to 
the  results  produced  or  the  phenomena  presented.     Thus 
we    speak    of   attraction,   repulsion,    cohesion,    friction, 
moving  forces,  accelerating  forces,  resisting  forces,  con- 
stant forces,  variable  forces,  muscular  forces,  vital  forces, 
etc. ;   but  they  are  all  alike  in  the  essential  quality,  that 
they  are  equivalent  to  a  pull  or  &push. 

31.  Measure   of  Force. — We  shall   assume   that   the 


131-33.]  FORCE.  17 

standard  pound  Avoirdupois  is  the  measure  of  a  unit  of 
force,  and  hence,  that  any  force  is  a  certain  number  of 
pounds.  We  are  familiar  with  the  fact  that  forces  are 
measured  by  pounds.  Thus,  if  a  spring  balance  of  suffi- 
cient strength  is  placed  between  a  locomotive  and  a  train 
of  cars,  it  will  indicate  the  pulling  force  of  the  locomotive, 
whether  the  train  remains  at  rest  or  is  in  motion.  A  push 
can  be  measured  in  the  same  manner. 

Forces  may  also  be  measured  by  considering  their  effect 
in  producing  motion  in  a  free  body,  as  will  be  shown  in 
Article  86. 

32.  Weight  is  a  measure  of  the  attractive  force  of 
gravity  upon  a  body. — Weight  varies  directly  as  gravity. 
It  has  been  found  that  the  same  body  will  weigh  more  in 
some  places  than  others.     It  will  weigh  a  trifle  less  on  the 
top  of  a  high  mountain  than  in  a  deep  valley,  and  if  it 
could  be  placed  where  there  was  no  attraction  it  would 
weigh  nothing.     It  is,  therefore,  necessary  to  designate 
some  place  where  a  body,  used  as  a  standard,  shall  be 
weighed  in  order  to  fix  the  standard  pound. 

33.  Standard  Measures. — The  standards   for  weights 
and  measures  are  established  by  legal  enactments.     The 
British  standard  for  one  pound  Avoirdupois  is  the  weight 
of  a  certain  .piece  of  platinum  kept  in  the  Exchequer  Office 
in  London. 

The  British  standard  yard  is  the  distance  between  two 
points  on  a  certain  metal  rod,  kept  in  the  Tower  of  Lon- 
don, when  the  temperature  of  the  whole  bar  is  60°  F. 
(=  15°.5C.).  The  foot  is  declared  to  be  one-third  of  the  yard. 

The  United  States  standards  were  copied  from  the 
British  standards;  but  it  has  since  been  found  that,  on 
account  of  errors  in  measurement,  the  British  yard  is  a 
trifle  shorter  than  the  American. 


18  KINETICS.  [33.] 

The  French  metre  is  the  distance  between  two  points  on 
a  certain  bar  kept  for  the  purpose,  and  is  nearly  10000<000  of 
the  length  of  a  meridian  measured  from  the  equator  to  the 
north  pole. 

The  relation  between  these  standards  and  certain  definite 
quantities  furnished  by  nature,  has  been  determined  with 
great  care.  For  instance,  after  Capt.  Ivater  determined 
very  accurately  the  length  of  a  pendulum  which  would 
vibrate  once  a  second  at  London,  compared  with  the  stand- 
ard then  in  use,  it  was  declared  by  Parliament  (5  Geo.  IV.), 
that  "  the  yard  shall  contain  36  parts  of  the  39.1393  parts 
into  which  that  pendulum  may  be  divided  which  vibrates 
seconds  of  mean  time  in  the  latitude  of  London,  in  vacuo, 
at  the  level  of  the  sea,  at  temperature  62°  F."  After  the 
standard  was  destroyed  by  fire,  the  commissioners  who 
examined  the  subject  reported  that  several  reductions  of 
the  pendulum  experiments  were  doubtful  or  erroneous, 
and,  accordingly,  the  distance  between  the  marks  on  a 
metallic  bar  was  adopted  as  the  standard,  and  the  above 
ratio  was  discarded. 

Similarly,  the  metre  was  originally  declared  to  be 
10,000,000  of  the  arc  of  the  meridian  measured  from  the  equa- 
tor to  the  north  pole,  and  the  French  government  expended 
large  sums  of  money  in  determining  this  distance.  Cer- 
tain arcs  of  the  meridian  were  measured  with  great  care, 
and  from  them  the  distance  was  computed.  Afterwards 
it  was  ascertained  that  there  had  been  errors  in  the  deter- 
mination, and  that  the  distance  was  not  the  same  on  all 
meridians,  for  the  equator  is  not  an  exact  circle  ;  but  the 
length  of  the  metre  was  not  changed,  and  hence  the 
declared  ratio  became  obsolete. 

Attempts  have  been  made  to  establish  a  legal  pound  by 
declaring  that  a  cubic  foot  of  water  at  its  maximum 


[34,35.]  FORCE.  19 

volume  shall  weigh  a  given  amount  (nearly  62^  pounds)  ; 
but  there  is  a  question  in  regard  to  the  temperature  of  the 
water  at  the  maximum  volume,  and  different  experimenters 
do  not  agree  as  to  the  weight.  Hence,  practically  and 
actually,  the  weight  of  a  certain  piece  of  metal  remains 
as  the  standard.  (See  The  Metric  System,  by  F.  A.  P. 
Barnard,  LL.D.,  New  York,  1872.)  The  Avoirdupois 
pound  is  7,000  grains.  For  the  relation  between  the 
English  and  French  measures  see  table  in  the  Appendix. 
The  French  metre  is  equivalent  to  39.37079  British  inches, 
or  39.368505  American  inches. 

The  unit  of  time  in  common  use  is  a  second  of  mean 
solar  time.  The  time  from  any  meridian  passage  of  the 
sun  to  the  following  one  is  called  a  solar  day,  or  simply  a 
day.  These  days  are  of  unequal  length,  caused  by  the 
variable  motion  of  the  earth  along  its  orbit,  while  the 
time  of  the  revolution  of  the  earth  on  its  axis  is  con- 
stant. Therefore,  the  average  length  of  the  day  for  an 
entire  year  is  used,  arid  called  24  hours,  from  which  the 
minute  and  second  are  easily  found.  The  second  is  the 
86,400th  part  of  a  mean  solar  day. 

34.  A  force  may  be  represented  by  a  straight  line. 
Thus,  in  Fig.  12,  the  magnitude  of  the  force  may  be 
represented  by  the  length  of  the  line 

AB,  the  point  of  application,  by  the     -^ ^-  •& 

point  A,  and  its  direction  of  action,  by  Fl°*  **• 

the  arrow-head,  indicating  that  it  acts  from  A  towards  B. 

35.  The  point  of  application  of  a  force  may  be  con- 
sidered as  at  any  point  in  its  line  of  action.     Thus,  if  a 
body  is  pushed  with  a  rod,  the  point  of  application  of  the 
force  may  be  considered  as  at  any  point  along  the  line  of 
the  rod ;    and,  if  the  rod  were  prolonged  through,  and 
beyond  the  body,  the  point  of  application  may  be  consid- 


20  KINETICS.  [36.] 

ered  as  at  any  point  on  the  prolonged  part.  The  effect 
upon  the  body  will  be  the  same,  whether  we  consider  the 
point  of  application  of  the  force  at  one  point  or  another 
of  the  rod. 

Inertia. 

36.  The  Inertia  of  matter  means  its  passiveness  •  thai 
is,  its  inability  to  change  its  own  condition  in  regard  to 
rest  or  motion.  If  a  body  be  at  rest,  it  has  no  power  to  put 
itself  in  motion,  or,  if  in  motion,  it  has  no  power  of  itself 
to  change  its  rate  of  motion.  It  implies  that  every  change 
of  motion  is  due  to  an  external  cause. 

A  body  not  constrained  by  other  bodies,  or  in  other 
words,  perfectly  free  to  move,  is  perfectly  sensitive  to 
the  action  of  a  force,  so  that  the  smallest  force  would 
move  the  largest  body.  The  mqpn,  earth,  and  other 
planets  are  in  this  condition,  and  each  yields  constantly  to 
the  action  of  all  the  others  upon  it.  It  is  said,  and  with 
good  reason,  that  every  step  which  a  man  takes  upon  the 
earth  changes  the  centre  of  gravity  of  the  earth. 

Inertia  is  not  a  force.  But  on  account  of  the  perfect 
passiveness  of  matter,  if  a  force  act  upon  a  free  body,  the 
effect  of  the  force  will  be  stored  in  the  body  ;  and  the  body, 
in  being  brought  to  rest  by  a  resistance,  will  produce  the 
same  effect  as  that  which  was  stored  in  it. 

The  exact  relations  which  exist  between  the  motion  of 
bodies  and  the  force  which  produces  it,  were  established 
only  after  a  long  series  of  observations,  experiments  and 
deductions.  It  is  difficult  to  prove  them  by  direct  experi- 
ments, for  it  is  difficult  to  realize  the  condition  of  a  per- 
fectly free  body.  Every  body  is  liable  to  be  resisted  by 
friction,  or  the  air,  or  other  medium  ;  and  if  the  body  be 
placed  in  a  vacuum,  the  range  of  its  motion  will  be  limited 


[37-39.]  MOLECULAR    MOTIONS.  21 

If  a  body  be  thrown  into  the  air,  it  not  only  meets  with 
the  resistance  of  the  air,  but  will  be  constantly  under  the 
action  of  the  force  of  gravity.  Delicate  experiments  have, 
however,  confirmed  all  the  fundamental  principles  of  mo- 
tion. Their  truth  is  also  shown  from  the  fact  that  all  prob 
lems  pertaining  to  the  motion  of  bodies,  not  only  on  the 
earth  but  in  the  solar  system,  solved  in  accordance  with 
these  principles,  give  results  which  agree  with  the  results 
of  observation.  The  times  and  places  of  eclipses  are  pre- 
dicted, and  the  positions  of  the  planets  are  foretold,  by 
means  of  formulas  which  grow  out  of  these  fundamental 
principles.  No  truths  in  science  have  become  more  firmly 

established. 

tt*~.  -    /Wj.H-,  — 1_, 

Molecular  Motions — Definitions. 

37.  Molecular  Motions. — We  have  thus  far  spoken  of 
the  motion  of  bodies,  but  a  close  examination  shows  that 
\h^  particles  of  a  body  may  have  a  motion  in  reference  to 
each  other.     Thus,  when  a  tuning-fork  or  a  bell  is  struck 
the   particles   are   put  into  a  rapid   vibratory  motion, 
producing  sound,  which  is  transmitted  to  the  ear  by  the 
vibrations  of  the   air.     Heat    expands  bodies,  an   effect 
which  must  be  caused  by  the  separation  of  the  particles  of 
the  body.     It  is  believed  that  the  molecules  or  atoms  of 
every  solid  are  made  to  vibrate  when  it  is  struck. 

38.  Mechanics  is  the  science  which  treats  of  the  action 
of  forces. — It  investigates   the  laws  which  govern   the 
action  of  forces ;    the  conditions  of  the  equilibrium  of 
bodies  ;  the  motion  of  bodies  under  the  action  of  forces ; 
the  flow  of  liquids  and  gases ;  and  the  movement  of  the 
particles  of  bodies. 

39.  Kinetics  treats  of  the  movement  of  bodies  under 
the  action  of  forces. — This  branch  of  the   subject  has 


22  KINETICS.  [40-47.] 

usually  been  called  Dynamics,  a  term  which  more  properly , 
pertains  to  the  abstract  doctrine  of  forces,  and  hence,  as 
such,  would  include  a  portion  of  both  Kinetics  and  Statics. 

40.  Statics  treats  of  the  conditions  of  the  equilibrium 
of  solids. 

41.  Molar  mechanics  treats  of  the  action  of  forces  upon 
solids. 

42.  Molecular  mechanics  treats  of  the  movement  of 
the  particles  of  a  body. 

43.  Hydrostatics  treats  of  the  equilibrium  of  liquids. 

44.  Hydrodynamics  treats  of  the  movement  of  liquids. 

45.  Pneumatics  treats  of  the  laws  of  pressure  and 
movement  of  air  and  other  gaseous  bodies. 

46.  Thermodynamics  treats  of  the  mechanical  prop- 
erties of  heat. 

47.  Rotation  of  Bodies. — A  force  acting  upon  a  body 
may  produce  both  translation  and  rotation  at  the  same 
time.     Thus,  a  boy  strikes  a  stick  with  a  club,  and  if  he 
does  not  strike  it  directly  opposite  the  centre  it  will  rotate 
at  the  same  time  that  it  moves  forward.     The  solution  of 
problems  involving  the  rotation  of  bodies  are  generally 
more  difficult  than  those  involving  translation  only.     If  all 
the  points  of  a  body  move  in  parallel  straight  lines,  the 
motion  is  that  of  simple  translation,  and  will  be  substan- 
tially the  same  as  if  we  consider  the  body  reduced  to  a 
mere  particle.     Hence,  in  discussing  the  subject  of  trans 
lation  it  is  common  to  speak  of  the  body  as  a  particle. 

EXERCISES. 

1.  If  a  body  be  suspended  by  a  very  long,  fine  string,  how  much  force 

will  be  required  to  move  it  sideways  ? 

2.  If  a  ship  could  float  on  water  without  any  resistance  from  the  water 

or  air,  how  much  force  would  be  required  to  move  it  ?    If  it  were 


T48-50.]          NEWTON'S  LAWS  OF  MOTION.  23 

* 

moving,  how  much  force  would  be  required  to  deflect  it  from  ita 
path? 

3.  Why  does  not  every  body  move  when  acted  upon  by  force  ? 

4.  If  four  spring  balances  are  connected  end  to  end,  and  a  man  pulls 

with  a  force  of  100  pounds  at  one  end  of  the  four,  what  will  be  the 
force  exerted  at  the  other  end,  and  what  will  each  balance  indicate  ? 
6.  If  a  constant  pull  of  500  pounds  is  exerted  at  one  end  of  a  rope, . 
which  is  attached  to  a  boat  at  the  other  end,  will  the  force  exerted 
at  the  other  end  be  500  pounds  when  the  boat  is  in  motion  ? 

6.  Suppose  that  two  heavy  sleds  are  placed  on  perfectly  smooth  ice  and 

connected  by  a  flexible  cord  ;  if  a  boy  draws  them  by  pulling  with 
a  constant  force  of  10  pounds  on  one  of  them  by  means  of  a  rope 
or  otherwise,  so  as  to  pull  both  sleds  in  the  same  direction,  will 
there  be  a  force  of  10  pounds  exerted  on  the  other  one  ?  That  is, 
will  the  tension  on  the  connecting  cord  be  10  pounds  ? 

(Perfectly  smooth  is  intended  to  imply  that  there  is  no  resistance 
between  the  sleds  and  ice.) 

7.  If,  in  the  preceding  exercise,  the  boy  ceases  to  pull,  but  all  the  other 

conditions  remain  the  same,  what  will  be  the  tension  upon  the 
connecting  cord  ? 

Newton's  Three  Laws  of  Motion. 

48.  Sir  Isaac  Newton  expressed  the  fundamental  prin- 
ciples of  the  relations  of  force  to  the  movement  of  bodies 
in  the  form  of  three  laws  or  axioms.     These  principles 
have  already  been  given  in  the  preceding  articles,  but 
these  laws  are  referred  to  so  frequently,  and  express  so 
clearly  and  correctly  the  fundamental  principles  of  the 
motion  of  bodies,  that  we  cannot  do  better  than  present 
them  in  this  place. 

49.  First  Law. — Every  body  continues  in  a  state  of 
rest  or  of  uniform  motion  in  a  straight  line,  unless  acted 
upon  ~by  a  force  which  compels  a  change. 

50.  Second  Law. —  Change  of  motion  is  in  proportion 
to  the  acting  force,  and  takes  place  in  the  direction  of  the 
straight  line  in  which  the  force  acts. 

Observe  that  the  resultant  motion  is  not  necessarily  iu 


24  KINETICS. 

4 

the  direction  of  action  of  the  deflecting  force,  b:.t  it  im- 
plies that  the  departure  from  the  line  of  motion  will  be 
proportional  to  the  deflecting  force.  To  illustrate :  if  a 
body  be  moving  due  south,  and  a  force  be  applied  which 
acts  in  a  due  easterly  direction,  like  the  wind,  for  instance, 
then  will  the  easterly  force  cause  it  to  move  east  of  the 
meridian  in  which  it  originally  moved,  and  the  amount  of 
this  departure  will  be  proportional  to  the  easterly  force, 
and  will  be  independent  of  the  velocity  in  the  southerly 
direction. 

51.  Third  Law. — To   every  action  there    is  always 
opposed  an  equal  and  contrary  reaction. 

That  is,  when  two  bodies  press  one  against  the  other, 
the  action  upon  one  is  exactly  equal  but  opposite  to  that 
upon  the  other.  If  one  presses  a  stone  with  his  finger,  the 
finger  is  pressed  equally  by  the  stone. 

The  term  motion  in  these  laws  means  more  than  velo- 
city ;  it  really  means  momentum,  a  term  which  will  be 
defined  hereafter. 

Resolved  Forces. 

52.  Parallelogram  of  Forces. — If  a  force  f[,  acting 
singly,  would  cause  a  particle  at  A  to  describe  the  line  AB 

in  a  time  t,  and  a  force  F%, 
also  acting  singly,  would 
cause  the  particle  to  pass  over 
AD  in  the  same  time ;  then, 
if  they  act  jointly  on  the 
same  particle,  they  will  cause 
it  to  describe  the  line  AC, 
Fia<  13<  the  diagonal  of  a  parallelo- 

gram, constructed  on  AB  and  AD,  in  the  same  time. 


f53-56.]  EESOLVED    FORCES.  25 

This  follows  directly  from  the  Second  Law.  For,  the 
force  Fz  will  cause  the  particle  to  reach  a  line  DO,  which 
is  parallel  to  AB,  in  the  same  time,  whether  JF[  acts  or 
not;  and  similarly  !*[  wjn  cause  it  to  reach  a  line  BO, 
parallel  to  AD,  in  the  same  time  ;  and  hence,  when  they 
act  jointly,  it  will  be  found  at  the  intersection  of  DO 
and  BO,  or  at  O,  at  the  end  of  the  time;  and,  as  it  must 
move  in  a  straight  line,  it  will  move  along  the  diago- 
nal AC. 

This  is  nearly  the  same  as  Article  14,  only  here  we  con- 
sider the  cause  of  the  motion,  while  there  the  velocities 
only  were  considered. 

53.  Resolved  Forces.  —  If  a  force  F  would  cause   a 
particle  to  describe  the  line  AC  uniformly  in  a  given 
time,  it  may  be  resolved  into  two  forces,  f[  and  F&  one  of 
which  will  cause  it  in  the  same  time  to  describe  the  side 
AJ5,  and  the  other,  AD  of  a  parallelogram,  of  which  A  O 
is  the  diagonal.     This  is  the  converse  of  the  preceding 
article.     The  result,  however,  is  indeterminate,  since  A  O 
may  be  the  diagonal  of  an  indefinite  number  of  parallelo- 
grams. 

54.  Components.  —  The  forces  JF[  and  F%  are  called 
component  forces  ;  and  F  is  the  resultant. 

55.  The  Resultant  of  two  or  more  forces  which  act 
upon  a  single  particle,  is  a  force  which  will  produce  the 
same  effect  as  all  the  other  forces  combined. 

56.  Relation  between  two  Forces   and  their  Re« 
sultant. 

Let 


a=  GAD-,  8=  CAB^ACD-, 

6  =  DAB  =  180°  -  ADO. 
2 


26  KINETICS. 

The  triangle  ADC  gives 

AD  DC 


[57,58..] 


AC 


or, 


sin  ACD~  sin  DA  C  ~  sin  AD  C ' 

77f  TTf  T7T 

JJ*)  J}-\  J1 


sn 


in  a       sin  6  ' 


sn 


57.  Rectangular  Components. — Let  the  components 
i  and  FI  make  a  right  angle  with  one  another;  and 

a  and  /9  be  the  same  as  in  the 
**  preceding  article.  Then  the 

triangle  A D  C  gives 

F%  —  F  cos  a  —  F  sin  /S. 

Squaring  and  adding,  observing 
Fio  c  that  sin  2a  +  cos  2a  =  1,  we  have 


58.  If  several  forces  act  along  the  same  line  upon  a 
body,  whether  in  the  same  or  opposite  directions,  their 
resultant  is  the  algebraic  sum  of  the  several  forces. 

Let  R  =  the  resultant,  then 


EXEKCISES. 

1.  When  a  body  is  thrown  horizontally  into  the  air,  why  does  it  fall  to- 

wards the  earth  ?    (See  the  SECOND  LAW.) 

2.  The  planets  are  free  bodies  in  space  ;  how  much  force  does  it  require 

to  deflect  them  from  their  course  ? 

8.  A  stone,  whose  weight  is  500  pounds,  rests  upon  another  stone  whose 
weight  is  2,000  pounds ;  what  is  the  reaction  of  the  latter  against 
the  former  ? 


[59,60.]  CONSTANT    FOECE.  27 

4.  -If  two  boats  of  equal  size,  offering  equal  resistances  to  movement  on 

water,  are  connected  by  a  rope,  and  a  man  in  one  of  the  boats  pulls 
on  the  rope,  drawing  the  boats  toward  one  another,  nt  what  point 
between  them  will  they  meet  ?  If  a  man  in  the  other  boat  also 
draws  in  the  rope,  at  what  point  will  they  meet  ? 

5.  If  a  person  is  on  ice,  which  moves  in  a  due  easterly  direction  at  the 

rate  of  1  mile  per  hour,  and  he  walks  on  it  at  the  rate  of  3  miles 
per  hour,  in  what  direction  must  he  travel  so  that  his  resultant 
course  shall  be  due  south  ? 

6.  If  a  ship  is  sailing  south-east  at  the  rate  of  8  miles  per  hour,  and  the 

tide  is  carrying  it  due  east  at  the  rate  of  3  miles  per  hour,  what  is 
its  actual  course,  and  its  velocity  ? 

7.  A  ship  is  sailing  due  south-east  at  the  rate  of  10  miles  per  hour  ;  what 

is  its  velocity  in  a  southerly  direction  ? 

8.  If  a  ball  is  placed  on  the  floor  of  a  railroad  car,  and  is  perfectly  free 

to  roll,  will  it  change  its  position  if  the  velocity  of  the  train  is 
gradually  increased  ?  Why  would  it  rush  toward  the  forward  end 
in  case  of  a  collision  with  another  train  ? 

9.  The  force  Fi  equals  20  pounds,  F2  equals  30  pounds,  and  the  angle 

between  them  is  60  degrees ;  required  the  value  of  their  resultant 
and  the  angle  between  it  and  the  forces  FI  and  F3  respectively. 


Constant  Force. 

59.  A  constant  force  is  one  which  acts  with  a  constant 
intensity. 

An  incessant  force  is  one  which  acts  constantly,  but  with 
a  variable  intensity. 

60.  A  constant  force  applied  to  a  free  body,  and  acting 
along  the  line  of  motion,  produces  a  constant  acceleration. 

For,  according  to  the  FIEST  LAW,  if  no  force  be  applied 
the  velocity  will  remain  constant,  and,  according  to  the 
second  law,  the  change  of  velocity  will  be  proportional  to 
the  force ;  and  as  the  force  remains  constant  the  change 
of  velocity  must  be  the  same  for  each  unit  of  time,  that  is, 
constant. 

If  the  force  be  applied  so  as  to  produce  an  increase  of 


28  KINETICS.  [61,62.] 

velocity,  the  acceleration  will  be  positive;  but  if  it  causes 
a  decrease  of  velocity,  the  acceleration  will  be  negative. 
In  the  former  case  the  velocity  would  increase  indefinitely, 
but  in  the  latter  case  the  body  might  be  brought  to  a  state 
of  rest.  For  instance,  a  body  thrown  horizontally  on  ice, 
or  a  train  of  cars,  moving  after  the  locomotive  is  detached, 
would  be  brought  to-  rest  by  friction.  An  active  force, 
operating  to  destroy  the  velocity,  may,  by  its  continued 
action,  produce  a  velocity  in  an  opposite  direction.  Thus, 
when  a  train  of  cars  is  in  motion,  the  locomotive  may  be 
reversed  and  push  against  the  train  to  stop  it,  and  by  con- 
tinuing its  action  may  finally  produce  a  velocity  in  an 
opposite  direction. 

61.  A  constant  moving  force   is   one   in   which   the 
resultant  of  all  the  forces  is  constant.     Thus,  when  a  loco- 
motive draws  a  train  of  cars,  it  may  exert  a  constant  mov- 
ing force  for  a  time  ;  that  is,  there  will  be  a  constant  force 
for  producing  motion ;  but,  as  the  velocity  increases,  the 
resistance  of  the  air  increases,  and  usually,  after  a  short 
time,  the  whole  power  of  the  locomotive  is  exerted  in 
overcoming  the  friction  and  resistance  of  the  air,  and  pro- 
duces- no  increase  of  velocity.     In  the  latter  case,  although 
the  locomotive  may  exert  a  constant  force,  it  is  not  called 
a  constant  moving  force,  for  the  pulling  force  of  the  loco- 
motive is  exactly  neutralized  by  the  resistances  of  the 
train. 

62.  The  movement  of  a  body  under  the  action  of  a 
constant  moving  force  is  illustrated  by  Fig.  9,  because 
the  force  produces  a  constant  acceleration.     Hence,  the 
formulas  of  Article  24  give  the  relations   between   the 
time,  space  and  velocity,  when  the  acceleration  is  known. 
The  line  of  action  of  the  force  is  supposed  to  be  in  tho 
direction  of  motion. 


[63-65.]  CONSTANT    FORCE.  29 

63.  Normal  action. — If  a  force  acts  constantly  normal 
to  the  path  described  by  the  body,  it  will  not  affect  the 
velocity.     Thus,  if  a  body  be  connected  to  a  point  by  a 
string,  so  that  the  body  will  describe  the  circumference  of 
a  circle,  the  tension  of  the  string  will  not  change  the 
velocity.     The  force  of  the  string  and  the  velocity  of  the 
body  may  both  remain  constant. 

64.  The  force  of  gravity  is  one  of  the  forces  of  nature. 
It  is  an  attractive  force,  tending  to  draw  bodies  towards 
each  other.     It  always  manifests  itself  wherever  there  is 
matter.     It  is  the  force  which  gives  weight  to  bodies,  and 
causes  unsupported  bodies  to  fall  to  the  earth.     It  holds 
the  planets  in  their  orbits.     It  acts  through  bodies  with- 
out being  diminished  in  its  intensity,  and  upon  the  most 
central  portions  of  a  body  with  the  same  intensity  as  if 
the  external  portions  were  removed. 

65.  The  Law  of  Universal  Gravitation  is  as  follows : 
Every  'particle  ATTRACTS  every  other  particle  in  the  DIRECT 
ratio  of  its  mass,  and  in  the  INVERSE  ratio  of  the  square 
of  its  distance. 

This  law  was  discovered  by  Sir  Isaac  Newton  in  1666, 
but,  on  account  of  an  erroneous  value  of  the  diameter  of 
the  earth  which  was  then  used,  he  was  not  able  to  prove 
it  at  that  time.     But  in  1682  it  was  found,  from  new 
measurements  of  an  arc  of  the  meridian,  that  the  correct 
diameter  was  about  TJT  greater  than  the  value  which  he 
had  previously  used  ;  and  with  this  value  he  fully  de- 
monstrated the  law. 
Let     M  =  the  mass  of  a  body  A  • 
in  =  the  mass  of  a  body  JB ; 
D  =  the  distance  of  the  body  A  from  a  body  C\ 
d  =  the  distance  of  the  body  B  from  a  body  C ; 
then 


30  KINETICS.  L66,  67.J 


attraction  of  A  upon  C       D*       dz   M 
attraction  of  J3  upon  C       m        in,    D* 

~d* 

66.  The  force  of  gravity  at  any  place  on  the  earth 
remains  constant.  —  This  is  shown  by  the  fact  that  the 
weight  of  a  body  is  always  the  same  at  the  same  place  ; 
also,  that  a  body  always  falls  the  same  distance  in  a 
vacuum  in  the  same  time  at  the  same  place. 

It  is  well  known,  however,  that  upheavals  are  taking 
place  in  some  parts  of  the  earth  and  depressions  in  others, 
and  these  doubtless  produce  exceedingly  slight  changes  in 
the  weight  of  a  body,  but  no  apparatus  at  the  present  day 
is  sufficiently  delicate  to  measure  these  changes,  if  they 
actually  exist  ;  hence,  we  may  say  that  the  force  of  gravity 
is  at  least  practically  constant  at  every  place. 

67.  Determination  of  the  acceleration  produced  by 
gravity  on  a  body  fatting  freely  in  a  vacuum.  —  We 
speak  of  a  vacuum  because  it  is  shown  experimentally 
that  all  bodies  fall  the  same  distance  in  the  same  time  in 
a  vacuum  ;  and  also  because  it  is  necessary  to  exclude  the 
resistance   of   the  air  in   determining  the  full  effect   of 
gravity.     This  is  a  problem  the  exact  solution  of  which 
involves  considerable  knowledge  of  mechanical  principles 
and  great  skill  in  making  observations.     These  principles, 
so  far  as  they  involve  the  use  of  the  pendulum,  will  be 
explained  hereafter,  but  in  this  place  we  can  only  describe 
the  process. 

By  means  of  Atwood's  machine  an  approximate  value 
of  the  acceleration  may  be  determined. 

By  means  of  delicate  machinery,  and  a  refined  system 
of  making  observations,  the  space  and  time  may  be 
observed  directly. 


[68.]  CONSTANT    FORCE.  31 

But  the  most  reliable  method,  or,  at  least,  that  most 
commonly  used,  is  by  means  of  a  pendulum,  Any  body 
vibrating  on  an  axis,  under  the  action  of  the  force  of  gravity, 
is  called  a  pendulum,.  If  the  vibrating  body  has  percepti- 
ble size,  it  is  called  a  compound  pendulum.  A  simple 
pendulum  is  a  material  particle  suspended  on  a  line 
without  weight,  and  hence,  is  an  ideal  pendulum,  but  it 
has  a  real  mathematical  signification.  A  simple  pendu- 
lum may  always  be  found  which  will  vibrate  in  the  same 
time  as  a  compound  one. 

If  I  =  the  length  of  a  simple  pendulum  ;  * 

t  =  the  time  of  one  vibration  in  seconds  ; 
g  =  the  acceleration  due  to  gravity  in  a  vacuum ; 
TT  =  3.141592  =  the  ratio  of  the  diameter  of  a  circle 

to  its  circumference ; 
then 


from  which  we  find 


~"w  ff9 


To  determine  t,  the  number  of  oscillations  for  a  given 
time,  say  10  or  20  minutes,  is  observed,  and  this  number, 
divided  by  the  number  of  seconds  in  the  time,  gives  t. 
The  length  I  is  measured  directly.  These  quantities,  sub- 
stituted in  the  preceding  formula,  will  give  the  value  of  g, 
which  in  this  latitude  is  about  32£  feet.  In  this  way  the 
intensity  of  gravity  has  been  found  at  different  places  on 
the  earth's  surface. 

68.  The  intensity  of  gravity  varies  with  the  lati- 
tude.— By  means  of  the  experiments  indicated  in  the 
preceding  article,  it  has  been  found  that  the  force  of 


32  KINETICS.  [69,70| 

gravity  is  least  at  the  equator,  and  increases  with  the  lati- 
tude both  north  and  south  of  the  equator.  The  value  of  <* 
for  any  latitude  L  may  be  found  approximately  by  the 
formula 

g  =  32.1726  -  0.08238  cos  2  L ; 

but  no  formula  will  give  the  exact  relation  between  g  and  L. 
At  the  equator  L  =  0,  and  at  the  poles  L  =  90° ;  hence  we 
have 

at  the  equator  gQ  =  32.0902  feet,  and 
u  at  the  poles      g^  =  32.2549  feet. 

69.  Weight  variable. — According  to  the  preceding 
article,  a  body  will  weigh  less  on  the  equator  than  at  any 
other  place  on  the  surface  of  the  earth.  But 
^*^  this  difference  could  not  be  detected  by  a  com- 
mon beam  balance,  for,  a  diminution  of  the 
weight  at  one  end  of  the  beam  would  be  exactly 
the  same  as  that  at  the  other,  and  if  the  bodies 
balanced  at  one  place  on  the  surface  of  the 
earth  they  would  balance  at  every  other  place. 
The  difference,  however,  might  be  detected  by 
a  spring  balance.  Fig.  15,  for  the  more  the  body 
weighed  the  more  it  would  pull  the  index  down. 


/M\ 


FIG.  is.  If  a  body  be  elevated  one  mile  above  the  sur- 
face of  the  earth,  it  will,  according  to  Article 
65,  lose  about  ^  of  its  weight.  These  variations  being 
small,  we  may,  for  the  purposes  of  this  work,  consider  g  as 
constant,  and  equal  to  32^-  feet. 

70.  There  are  three  causes  for  the  variation  of 
gravity  on  the  surface  of  the  earth. — First,  the  earth  is 
an  oblate  spheroid,  the  axis  of  which  coincides  with  the 
axis  of  the  earth ;  and  those  bodies  on  the  equator,  bejng 
more  remote  from  the  centre  of  the  earth  than  those  at 


F71.] 


CONSTANT    FORCE. 


33 


liigher  latitudes  *  will  be  attracted  with  less  intensity ; 
second,  the  revolution  of  the  earth  on  its  axis  produces  a 
so-called  "  centrifugal  force  "  (to 
be   explained   hereafter),   which 
tends  to  throw  bodies  from  the 
surface,   thus   diminishing  their 
weight ;  and  third,  the  heteroge- 
neous character  of  the  substance 
of  the  earth. 

The  form  of  the  earth  is  sup- 
posed to  be  due  to  the  attraction 
of  the  particles  for  each   other 
and  to  the  centrifugal  force  caused  by  the  rotation  on  its 
axis  while  the  substance  of  the  earth  was  in  a  plastic  state. 

71.  The  atmosphere  resists  the  movement  of  bodies  in 
it ;  and  hence,  the  velocity  of  bodies  under  the  action  of 
any  force  is  less  than  it  would  be  in  a  vacuum.  The 
attraction  of  the  earth  being  the  same  on  each  particle  of 
a  body,  a  light  body  would  fall  as  rapidly  as  a  heavy  one 
if  there  were  no  resistances  to  their  movements ;  and  this 
is  confirmed  by  experiment,  by  letting  bodies  fall  in  a 
vacuum.  The  resistance  of  the  air  varies  with  the  surface 
against  which  it  acts,  but  in  falling  bodies  the  ability  to 
overcome  this  resistance  varies  as  the  weight  of  the  body ; 
hence,  heavy  bodies  fall  faster  than  light  ones  in  the  air 
But  the  velocities  of  heavy  bodies,  such  as  iron,  stone, 
brass,  etc.,  falling  100  to  200  feet,  do  not  differ  much  from 

*  As  determined  by  The  semi-polar  axis  is  and  the  equatorial  radios  is 

Bessel 20,853,662  feet ;  20,923,596  feet. 

Airy 20,853,810  feet ;  20,923,713  feet 

Clarke  20,853,429  feet ;  20,923,161  feet. 

The  equatorial  diameter  of  the  earth  is  about  26  miles  longer  than  its 
axis 

2* 


34  KINETICS.  f  78-75.) 

each  other;  and  for  compact  masses  of  such  materials 
falling  in  air  we  use  32|-  feet  for  g. 

formulas  for  Falling  Bodies. 

72.  Bodies  falling  from  rest.  —  The  acceleration  being 
constant,  we  have  only  to  substitute  its  value  in  the  equa- 
tions of  Article  24.  Making  ,f=ff  and  s  =•  h  in  those 
formulas,  we  have. 

*=>*;    .....   (i) 

....     (2) 
;          .        .        .     (3) 


v        /2/A 

=  -  =  [  —  I 

ff        \ff/ 


and  t  =  -  =  [  —  I  =  —  .       .        .     (4) 
ff        \ff/         v 

73.  If  a  body  is  projected  downward  with  a  velocity 
V0,  we  make/"  =  g  and  s  =  h  in  Article  25,  and  have 

v  —  gt  +  VQ.  •         •         •     (5) 

h  =  \gf  +  v0t.      .         .         .     (6) 

74.  If  a  body  is  projected  upward  with  a  velocity, 
the  acceleration  becomes  negative,  and  the  equations  of 
the  preceding  article  become 

v  =  v0  —  gt.          .        .        .     (7) 
h  =  vj-$g&.      .        .        .     (8) 

75.  Problems.  —  1.  If  a  body  is  projected  upward  with 
a  velocity  of  100  feet  per  second,  required  its  height  at 
the  end  of  2  seconds. 

Equation  (8)  gives 

h  =  100  x  2  -  i  x  32|  X  4  =  135|  feet. 
2.  In  the  preceding  problem  wJuit  will  be  the  height  at 
the  end  of  8  seconds  f 
We  have 


[75.]         FORMULAS    FOR    FALLING    BODIES.  35 


h  =  8  x  100  -  £  x  32£  x  82  =  -  229£  feet  ; 
that  is,  at  the  end  of  8  seconds,  the  body  will  be  229-J  feet 
below  the  starting  point. 

3.  In  Problem  1,  what  will  be  the  greatest  height  of 
ascent  ? 

When  it  is  at  the  greatest  height  v  will  be  zero  in  equa- 
tion (7)  ;  hence 

v0  =  gt. 

.'.  *  =  -°  =  3.1  +  sec. 

g 

and  this  in  equation  (8)  gives 

h  =  100  x  3.1  -  i  x  32£  x  (3.1)8  =  156.4  feet. 

4.  If  a  body  is  projected  upward  with  a  velocity  of  200 
feet  per  second,  required  its  height  when  the  velocity  is 
100  feet  per  second. 

From  equation  (7),  we  find  for  the  time 

200  -  100       600 
-32T       rl93 
which  substituted  in  equation  (S)  gives 


EXAMPLES. 

/ 

1.  A  body  falls  from  rest  through  a  height  of  100  feet; 

required  its  final  velocity.     (Let  g  =  32  J  feet.) 

2.  A  body  falls  from  rest  and  acquires  a  velocity  of  300 

feet  ;  required  the  time. 

3.  A  body  is  projected  upward  with  a  velocity  of  100  feet 

per  second'  what  will  be   the  greatest  height  of 
ascent  ? 


36  KINETICS.  \15.\ 

4.  If  g  =  32£  feet  per  second,  what  will  be  the  accelera- 

tion per  minute. 

5.  A  metre  is  3.28  feet  (nearly)  ;  if  the  unit  of  time  were 

2  seconds  what  would  be  the  accelerator  ? 

6.  If  a  body  is  projected  downward  with  a  velocity  of  25 

feet  per  second,  what  will  be  the  velocity  after  it  haa 
fallen  120  feet  3 

7.  In  the  preceding  example,  what  will  be  the  time  of 

descending  150  feet  ? 

8.  At  the  instant  a  body  is  dropped  from  a   point   A, 

another  body  is  projected  upward  from  a  point  B, 
vertically  under  A,  arid  they  meet  at  the  middle  of 
AB\  required  the  velocity  of  projection  from  B. 

Ans.  (AB  x  </)*. 

9.  A  body  is  let  fall  into  a  well,  and  4  seconds  afterward 

it  is  heard  to  strike  the  water ;  if  the  velocity  of 
sound  is  1130  feet  per  second,  required  the  depth  of 
the  well. 

Ans.  231  feet. 

10.  A  body  is  projected  downward  from  a  point  A  with  a 

velocity  of  v  feet  per  second,  and  another  body  is 
projected  upward  from  a  point  a  feet  below  the 
former,  with  a  velocity  of  V  feet  per  second ; 
required  their  point  of  meeting. 

EXERCISES. 

1.  If  a  boy  draws  a  load  on  a  sled   with  an  increasing  velocity,  will  he 

exert  any  more  force  than  if  he  draws  it  at  a  uniform  rate  ? 

2.  Why  will  a  body  fall  more  rapidly  at  the  foot  of  a  mountain  than  at 

its  top  ? 

3.  Why  may  some  light  bodies  fall  more  rapidly  at  the  top  of  a  moun- 

tain than  at  its  foot  ? 

4.  If  a  body,  whose  weight  is  5  Ibs.,  attracts  a  particle  at  a  distance  of 

5  feet  with  a  force   of  TOTS  of  an  ounce,    with  what  force  wiT 


[76.]  ATTRACTION    OP    SHELLS.  37 

another  body,  whose  weight  is  25  Ibs.,  attract  the  same  particle  at 
a  distance  of  15  feet  ? 

5.  Which  will  vibrate  in  a  shorter  time,  a  pendulum  10  inches  long  or 

one  15  inches  long  ? 

6.  If  a  pendulum  vibrates  once  each  second  at  New  York,  will  the  time 

of  a  vibration  of  the  same  pendulum  be  more  or  less  than  a  second 
at  the  equator  ? 

7.  If  a  merchant  weighs  iron  in  New  York  and  sends  it  to  some  port 

near  the  equator  ;  will  it  gain  cr  lose  in  weight  if  it  be  weighed  in 
both  places  with  the  same  beam  scales  ?  Will  it  gain  or  lose  if 
weighed  with  the  same  spring  balances  ? 

8.  What  is  the  value  of  g  (32^  feet)  in  metres  per  second  ? 

Attraction  of  Homogeneous  Shells. 

76.  Problem.  —  The  attraction  of  a  perfectly  homoge- 
neous ,  spherical  shell  is  the  same  upon  a  particle  placed 
anywhere  within  it. 

Let  ABCDE\>Q  a  section  of  an  indefinitely  thin  spheri- 
cal shell,  and  0  any  point  within  it.  Draw  lines  BOD 
and  AO  (7  through  the  point  O,  mak- 
ing an  indefinitely  small  angle  with 
each  other,  and  consider  A  OB  and 
COD  as  two  cones  having  their  verti- 
ces at  O,  and  their  bases  AB  and 
CD  in  the  surface  of  the  sphere.  The 
quantities  of  matter  in  each  will  be  Fia 

directly  as  their  bases,  but  at  the  limit 
the  triangles  are  similar,  and  the  bases  will  be  as  A  Oz  to 
OC2.     The  attraction  of  each  varies  as  the  quantity  of 
matter  and  inversely  as  the  square  of  the  distance  from  0  ; 
(see  Article  65)  that  is, 

~ 


Attraction  of  AB  (AOf 

Attraction  of  CD 


38  KINETICS.  [77,  78.1 

hence,  the  attraction  of  the  bases  will  equal  one  another^ 
and,  being  in  opposite  directions,  will  neutralize  one 
another's  effect.  Similarly,  conceive  that  the  shell  con- 
stitutes the  bases  of  an  indefinitely  large  number  of  cones, 
then,  according  to  the  above  reasoning,  the  attraction  cf 
all  the  matter  on  one  side  of  any  straight  line  drawn 
through  O  will  exactly  neutralize  that  on  the  other  side. 
The  same  may  be  proved  for  any  other  point. 

77.  Problem. — If  the  earth  were  a  homogeneous,  hollow 
sphere  of  uniform  thickness,  a  body  placed  at  any  point 
within  the  hollow  would  remain  at  rent. 

For,  according  to  the  preceding  article,  the  attraction  of 
any  of  the  spherical  shells  of  infinitesimal  thickness  upon 
any  point  within  it  will  be  zero,  and 
hence,  the  effect  of  all  of  them  upon 
the  same  point  will  be  neutralized. 

It  follows  from  this  that  if  the  earth 
were  solid  and  composed  of  homogene- 
ous, concentric  shells,  though  they  var- 
ied according  to  any  law  from  the  centre 
to  the  surface,  the  resultant  attraction 
upon  a  particle  at  the  centre  of  the  earth  would  be  zero. 

78.  Problem. — If  the  earth  were  a  homogeneous  solid 
sphere,  the  resultant  attraction  upon  any  point  within  it 
would  vary  directly  as  its  distance  from  the  centre  of  the 
earth. 

Suppose  that  a  particle  is  at  a  distance  x  from  the  cen- 
tre ;  then,  according  to  the  preceding  problem,  the  result- 
ant attraction  of  the  shell  outside  of  x  will  be  zero ;  and 
that  portion  of  the  sphere  whose  radius  is  x  will  attract 
directly  as  its  quantity  of  matter  and  inversely  as  the 
square  of  the  distance  of  the  particle  from  the  centre  of 
the  sphere. 


l?9-81.]  WEIGHT  —  MASS  —  DENSITY.  39 

.    The  quantity  of  matter  will  be  directlj  as  the  volume, 
or  as 


and  hence,  the  attraction  will  vary  as 

£  TT  a?  -T-  a?2,  or  as  -|  TT  x  ; 

that  is,  directly  as  the  distance  of  the  particle  from  the 
centre. 

Weight  —  Mass  —  Density. 

79.  Weight.  —  The  weight  of  a  body  has  already  been 
defined,  in  Article  32,  as  a  measure  of  the  attractive  force 
of  the  earth  upon  the  body. 

Weight  is  not  essential  to  matter.  According  to  Arti- 
cle 76,  if  a  body  were  placed  at  the  centre  of  the  earth  it 
would  weigh  nothing.  Similarly,  if  placed  at  a  certain 
point  between  the  earth  and  moon  it  would  lose  its  weight. 
According  to  the  preceding  articles  its  weight  depends  upon 
its  position  ;  or  more  specifically,  upon  the  attraction  of  the 
earth  upon  it. 

80.  Mass  is  a  term  used  to  express  quantity  of  matter. 
—  This,  as  we  have  seen,  is  independent  of  the  weight  of 
the  body  ;  in  other  words,  it  is  constant  for  the  same  body. 
The  ratio  of  the  weights  of  two  bodies  in  vacuo,  deter- 
mined at  the  same  time,  whether  on  the  equator,  the  top 
of  a  mountain,  within  the  earth,  or  at  any  place  in  the 
universe,  is  constant.     Hence,  if  two  bodies,  one  weighing 
2  pounds  and  the  other  10  pounds,  be  balanced  on  a  lever 
at  any  place,  they  would  remain  balanced  if  taken  to  a 
place  where  the  bodies  weighed  1  and  5  pounds  respect- 
ively, or  4  and  20  pounds  respectively. 

81.  Measure  of  Mass.  —  The  mass  of  a  lody  equals  its 


40  KINETICS.  fSl.j 

weight  at  any  place  divided  by  the  acceleration  due  to. 
gravity  at  the  same  place. 

It  must  be  measured  in  such  a  way  as  to  give  the  same 
value  wherever  determined.  As  already  shown,  the  weight 
varies  directly  as  the  force  of  gravity,  and  the  acceleration 
due  to  gravity  also  varies  as  the  same  force  ;  hence,  the 
ratio  of  the  weight  to  the  acceleration  will  be  constant  for 
all  places,  both  being  determined  at  the  same  place. 
If 

M  =  the  mass  of  a  body  ; 
W  —  the  weight  of  the  body  at  any  place,  and 
g  =  the  acceleration  due  to  the  force  of  gravity 

at  the  same  place  ; 
then, 

i/-       W 

M«  F' 

which  may  be  put  in  the  form  of  an  equation  by  intro- 
ducing a  constant  c,  hence, 

W 
M  =  C  -  . 

g 

Since  G  is  arbitrary,  the  unit  of  mass  may  be  so  chosen 
that  c  will  be  unity,  in  which  case  we  have 


which  is  the  value  used  in  Mechanics. 

The  weight  in  this  case  must  be  determined  by  a  spring 
balance,  or  its  equivalent,  which  must  weigh  correctly  a 
standard  pound,  or  multiples  thereof. 

For  ordinary  practical  purposes  it  is  only  necessary  to 
divide  the  weight  of  a  body,  determined  at  any  place  on 
the  earth  with  a  pair  of  good  scales,  by 


[82-83.]  WEIGHT  — MASS  — DENSITY.  41 

The  mass  of  a  body  might  also  be  found  by  weighing  it 
at  any  place  with  a  pair  of  beam  scales,  using  as  weights 
standard  units  of  mass ;  that  is,  32£  standard  pounds  of 
weight. 

82.  The  mass  of  a  body  is  the  number  of  pounds  that  a 
body  would  weigh  at  that  place  where  the  acceleration  due 
to  gravity  is  one  foot  per  second,  the  weight  being  deter- 
mined by  a  standard  spring  balance. 

In  the  last  equation  of  Article  81  make  g  =  1,  and  let- 
ting Wi  be  the  corresponding  weight,  we  have 

M=Wi. 

One  such  place  is  in  the  earth,  at  about  -^  of  the  radius 
of  the  earth  from  the  centre,  or  less  than.  125  miles  from 
the  centre  of  the  earth.  Another  point  is  about  22,000 
miles  from  the  centre ;  for  if  x  be  the  distance,  and  the 
radius  of  the  earth  be  called  4,000  miles,  we  have,  accord- 
ing to  Article  65, 

(4000)2  :  a?  : :  1  :  32| ; 
.-.  x  =  22,686  miles. 

Since  the  weight  of  the  same  body  varies  directly  as  the 
force  of  gravity,  we  have 

M pounds  :  W  pounds  :  :  1  foot :  g  feet ; 

w      W 

.'.  M  = . 

ff 

In  this  expression  g  may  be  considered  as  an  abstract 
number,  being  the  ratio  of  the  acceleration  due  to  gravity 
at  two  different  places. 

83.  The  Unit  of  Mass  is  a  body  which  weighs 
standard  pounds. 

For,  if  M  =  1  in  the  preceding  article,  we  have 
W  pounds  =  g  x  1  pound  =  32%  pounds. 


4:2  KINETICS.  [84,85.1 

The  term  pounds  is  used  in  a  double  seuse.  We  use 
pounds  of  weight  and  pounds  of  mass,  but  no  ambiguity 
arises  on  account  of  it,  for  the  language  of  the  problem 
will  always  determine  which  js  referred  to. 

84.  Analytical  expression  for  Weight.  —  From  the 
last  equation  of  Article  82  we  have 


an  expression  which  is  useful  in  the  solution  of  many 
problems. 

85.  Density  relates  to  the  compactness  of  matter.  If 
the  mass  of  a  body  be  uniform  throughout  the  volume,  the 
density  is  the  mass  of  a  unit  of  volume. 

Let 

M  =  the  mass  of  a  homogeneous  body, 
V=  the  volume  of  the  body, 
D  =  its  density  ; 
then, 

D  =  y.    .:M=DV. 

If  V=  1,  then 

X>  =  M. 

Substitute  the  value  of  M  from  the  last  equation  of 
Article  82,  and  we  have 


If  the  density  be  variable  the  density  at  any  point  of  a 
body  will  equal  the  mass  of  a  unit  of  volume  having  the 
same  density  throughout  the  unit  as  that  at  the  point  of 
the  body  considered. 

The  unit  of  volume  may  be  taken  as  a  cubic  inch,  foot, 
or  any  other  standard  measure. 


[85.]  MEASURE    OF    FORCE.  43 

xl 

. 

EXEECISE8. 

1.  If  the  earth  were  a  homogeneous  sphere,  and  a  body  on  its  surface 

weighed  10  Ibs. ,  what  would  be  its  weight  if  placed  at  a  point 
half  way  between  the  surface  and  centre  ? 

2.  If  a  body  weighs  10  Ibs.  on  the  surface  of  the  earth,  what  would  it 

weigh  if  elevated  to  a  point  above  the  surface  equal  to  the  radius 
of  the  earth  ? 

3.  If  the  earth  were  a  homogeneous  spherical  shell  of  finite  thickness, 

what  would  be  the  weight  of  a  body  placed  anywhere  within  the 
hollow  ? 

(The  earth  is  referred  to  in  these  questions  because  the  condi- 
tions could  not  be  realized  by  a  hollow  sphere  on  the  surface  ;  for 
gravity  would  exert  its  full  force  on  a  body  placed  in  the  hollow 
of  such  a  sphere.) 

4.  In  the  preceding  example,  if  a  body  weighed  10  Ibs.  on  the  surface 

of  the  earth,  at  what  place  in  the  shell  must  it  be  placed  that  its 
weight  shall  be  5  Ibs.  ? 

5.  If  the  earth  were  a  homogeneous  shell,  and  a  body  were  dropped 

from  the  surface  into  the  hollow,  would  the  motion  be  accelerated 
or  not  as  it  passes  through  the  shell  ?  And  as  it  passed  across 
the  hollow  would  its  motion  be  accelerated  or  not,  no  allowance 
being  made  for  the  resistance  of  the  air  ? 

6.  If  a  person  were  placed  in  the  hollow  described  in  the  preceding 

question,  and  should  jump  from  one  side  toward  the  centre  of 
the  sphere,  where  would  he  stop  ?  Could  he  stop  at  the  centre  if 
he  desired  to  ? 

7.  If  a  person  were  placed  at  one  extremity  of  a  diameter  of  the  hollow 

referred  to  in  exercise  5,  and  a  ball  of  equal  mass  placed  at  the 
other  extremity,  and  the  person  should  pull  on  the  body  by  means 
of  a  string,  where  would  they  meet  ?  If  he  pulls  for  an  instant 
and  then  ceases  to  pull,  will  they  meet  ?  If  he  pulls  with  a  constant 
force  until  they  meet,  will  their  acceleration  be  uniform  or 
variable  ? 

8.  In  the  preceding  question,  if  the  ball  has  half  the  mass  of  the  person, 

which  will  move  faster,  the  ball  or  the  person  ? 

0.  If  a  person  were  placed  at  the  centre  of  the  hollow  sphere  of  exer- 
cise 5,  and  not  able  to  reach  anything,  could  he  move  away  from 
the  centre  by  his  own  exertions  ?  -If  he  had  a  ball  and  should 
throw  it  away,  would  the  person  move  away  from  the  centre? 


44  KINETICS.  [86,87.] 

(He  could  not  drop  the  ball,  for  it  would  not  move  in  any  direction 
unless  started.)  What  would  be  their  relative  directions  of  motion  1 
Would  they  move  in  straight  or  curved  lines  ? 

10.  If  a  body  weighs  100  standard  pounds,  how  many  pounds  of  mass 

does  it  contain  ? 

11.  If  a  body  whose  volume  is  2  cubic  feet  weighs  200  Ibs.,  what  is 

its  density  ? 

12.  If  a  body  weighs  5  kilogrammes,  what  is  the  mass  in  pounds  ? 

13.  If  the  weight  of  a  body  whose  volume  is  1  cubic  metre  is  3  kilo- 

grammes, what  is  the  density  in  pounds  per  cubic  foot  ? 

14.  If  a  hole  were  made  through  the  centre  of  the  earth  from  surface 

to  surface,  and  a  ball  were  dropped  into  it,  would  it  stop  at  the 
centre  ?  Where  would  it  stop  if  the  hole  were  a  vacuum  ?  At 
what  point  would  it  move  with  the  greatest  velocity  ? 


Dynamic  Measure  of  Force. 

86.  Value  of  a  Moving  Force.  —  Forces  are  compared 
by  their  effects.  If  a  force  F,  acting  as  a  constant  pull 
or  push  on  a  body  perfectly  free  to  move  in  the  direction 
of  action  of  the  force,  produces  an  acceleration  f,  and 
gravity  acting  on  the  same  body  with  a  constant  force  7F", 
equal  to  the  weight  of  the  body,  produces  an  acceleration 
0,  we  have 

F:  W::f:g 


that  is,  a  constant  moving  force  of  F  pounds  equals  the 
product  of  the  mass  into  the  acceleration  in  feet  per 
second. 

From  the  last  equation  we  have 
F        F 
f^'M'-''  Wg' 

87.  Effective  Force.  —  Only  a  portion  of  the  forces 
which  act  upon  a  body,  may  be  effective  in  moving  it,  the 


[88,89.]  MEASURE   OF    FORCE.  45 

others  neutralizing  each  other.  Thus,  when  a  locomotive 
draws  a  train  of  cars,  a  portion  of  the  pulling  force  is 
directly  neutralized  by  the  resistance  of  the  air,  friction 
on  the  track,  and  other  resistances  of  the  train.  If  the* 
pulling  force  exceeds  the  resistances,  the  excess  will  be  the 
effective  pulling  force.  When  the  resistances  equal  the 
pulling  force  the  motion  will  be  uniform. 

88.  Remark. — To  find  the  space  passed  over  by  a  body, 
and  the  velocity  produced  in  a  given  time  by  the  action  of 
a  constant  effective  force,  find  the  value  of  f  from  Article 
86,  and  substitute  its  value  in  the  equations  of  Article  24. 

89.  Problems. — 1.  If  a  piston  be  driven  a  portion  of 
the  length  of  a  cylinder  by  a  constant  steam  pressure,  at 
what  point  must  the  pressure  be 

instantly  reversed  so  that  the  full 
stroke  shall  just  equal  the  length 
of  the  cylinder,  the  cylinder  being 
horizontal,  and  the  piston  moving 
without  friction  ?  Fjo  19 

At  the   middle  of    the    stroke. 

Whatever  velocity  is  generated  by  the  action  through  one 
half  of  the  stroke  will  become  neutralized  by  the  same 
pressure  acting  in  the  opposite  direction  through  the 
remaining  half. 

2.  In  the  jweceding  example,  what  will  be  the  velocity 
at  the  centre  of  the  cylinder  f 

Let      F  =  the  total  pressure  of  the  steam, 
W  =  the  weight  of  the  piston, 

s  =  one-half  the  length  of  the  cylinder ; 
then,  from  Article  86,  we  have 

F 

•*•      . 

—  ~W  y  > 


46  KINETICS.  [89.] 

and,  if  the  piston  starts  at  one  end,  we  substitute  thia 
value  in  equation  (3)  of  Article  24,  and  find 


3.  A.  body  whose  weight  is  W  is  moved  horizontally  on 
a  frictionless  surface  by  the  pull  of  a  constant  force  F; 

required  the  space  passed  over 
in  a  time  t,  and  the  velocity 

FIO.  so.  Here  we  have, 

f=  — 
hence,  from  Article  24,  we  find 

and  ^  " 


= 


4.  If  a  body,  whose*  weight  is  W  on  the  surface  of  the 
earth,  be  placed  in  the  hollow  sphere  described  in  Article 
76 ;  required  the  constant  putt  or  push  necessary  to  pro- 
duce a  velocity  v  in  time  t. 

From  Article  86  we  have 

f       F 
f--'--^9\ 

which,  substituted  in  equation  (1)  of  Article  24,  gives 

F 
'"Wfft> 

from  which  we  find 


[90.]  MEASURE   OF   FORCE.  47 

EXAMPLES. 

1.  If  a  piston  weighs  100  Ibs.,  and  the  constant  steam 
pressure  is  2,000  Ibs.,  what  will  be  the  velocity 
acquired  in  moving  over  12  inches  ? 

J5.  In  the  third  problem  above,  if  the  weight  of  the  body 
is  500  Ibs.,  and  the  constant  pulling  force  is  25  Ibs., 
required  the  space  over  which  the  body  will  be 
moved  from  rest  in  10  seconds. 

3.  In  the  same  problem,  if  the  constant  frictional  resist- 

ance is  10  Ibs.,  what  will  be  the  velocity  at  the  end 
of  100  feet? 

4.  In  the  fourth  problem  above,  if  a  body  weighs  100  Ibs. 

at  a  place  where  g  =  32  feet,  and  is  placed  in  a 
hollow  space  at  the  centre  of  the  earth ;  required 
the  pulling  force  .necessary  to  produce  a  velocity  of 
100  feet  in  10  seconds. 

90.  Problems. — 1.  Suppose  that  a  body  is  on  a  friction- 
Less  plane,  and  is  moved  horizontally  by  a,  weight  attached 
to  it  by  means  of  a  string  pass- 
ing over  a  pulley  •  required 
the  space  passed  over  by  each  J 
of  the  bodies  in  a  time  t,  there 
being  no  resistances  from  the 
pulley,  string  or  air.  ^  a 

'         P  =  the  effective  moving  force, 

P  +  W 

=  the  total  mass  moved  ; 

g 

hence,  from  Article  86,  we  have 

P  P 

f  —        ..  -     /T. 

•*  ~  P  +  Wff> 

g 


48  KINETICS.  [90.] 

and  this,  in  equation  (4)  of  Article  24-,  gives 


(REMAKK. — If  the  student  finds  it  difficult  to  distinguish  between 
the  moving  force  and  the  mass  moved,  let  him  imagine  the  whole 
system  placed  in  the  hollow  sphere  described  in  Article  76.  Then  both 
bodies,  P  and  TF",  will  be  destitute  of  weight,  and  no  motion  can  result 
from  the  action  of  P.  Now  conceive  that  a  string  is  attached  to  the 
body  P,  and  passed  through  the  shell  to  some  point  on  the  surface 
where  a  man  pulls  with  a  constant  force  of  P  pounds  ;  the  result  will 
be  the  same  as  that  given  in  the  preceding  problem,  for  the  man  will 
be  obliged  to  move  the  mass  of  both  bodies.  When  the  system  is  on 
the  earth,  gravity  pulls  with  a  force  of  P  pounds  on  the  body  P,  and 
nothing  in  the  direction  of  motion  of  FT;  hence  the  force  P  must  move 
both  P  and  W.) 

2.  Required  the  tension  of  the  string  in  the  preceding 
problem. 

Let  T  —  the  tension. 

Conceive  the  string  to  be  severed,  and  a  force  applied 
equal  to  the  tension  T,  pulling  upward  on  the 
body ;  then  will  the  effective  moving  force  fa 

(P  -  T)  pounds. 
The  mass  of  the  body  P,  will  be 

FIG.  82.  —  . 

hence,  according  to  Article  86,  we  have 


The  effective  pulling  force  on  TF,  Fig.  21,  is  the  tension 
of  the  Btring,  hence,  according  to  Article  86, 


[90.] 


MEASURE    OF    FORCE. 


Eliminating  f  from  these  equations  gives 

WP 


T  = 


W  +  P' 


3.  A  string  passes  over  a  wheel  and  has  a  weight  P 
attached  at  one  end  and  W  at  the  other  /  if  there  are  no 
resistances  from  the  string,  wheel,  or  air, 
and  the  string  is  devoid  of  weight,  required 
the  resulting  motion. 

The  effective  moving  force  F,  =  (  W—  P) 
pounds  / 

W+P 

the  mass  moved  = — ; 


hence,  from  Article  86, 


9 
,      TF-P 

**/: :  w+p  ff> 

and,  from  Article  24,  we  have  for  the  space 

W-P 


and  for  the  velocity, 


v  == 


W+P 

W-P 

W+P 


(These  problems  are  ideal,  since  they  discard  certain  elements,  such 
as  the  mass  of  the  pulley,  friction,  and  stiffness  of  the  cord.  These, 
however,  may  all  be  computed,  but  they  make  the  problem  too  compli- 
cated for  this  part  of  the  work.  The  chief  object  here  is  to  confine  the 
attention  to  the  relation  between  forces  and  the  motion  produced  upon 
masses. ) 


50  KINETICS.  [90.] 

4.  Find  the   tension  of  the  string  in  the  preceding 
problem. 

2PTF 


5.  A.  man,  whose  weight  is  TF,  stands  on  the  platform 
of  an  elevator  as  it  descends  the  vertical  shaft  of  a  mine  • 
if  the  platform  descends  with  a  uniform  acceleration  of 
%g,  show  that  his  pressure  upon  the  platform  is  £•  TF. 
What  will  it  be  if  the  platform  ascends  with  the  same 
uniform  acceleration  ? 

EXAMPLES. 

1.  In  Fig.  21,  if  P  =  TF,  how  far  will  the  bodies  move  in 

5  seconds  ? 

2.  In  Fig.  23,  if  TF  =  2P,  how  far  will  the  bodies  move 

in  5  seconds? 

3.  In  Fig.    23.      TF  =  50  pounds;    what  must   be   the 

weight  of  P  so  that  it  will  descend  10  ft.  in  5 
seconds  ?  What,  that  it  may  ascend  10  ft.  in  5 
seconds  ? 

4:.  In  Fig.  23,  if  P  =  5  ounces,  and  TF  —  4£  ounces, 
and  it  is  observed  that  P  descends  6.8  ft.  in  2 
seconds ;  required  the  value  of  g. 

5.  In  Fig.  23,  if  TF  =  10  pounds,  required  the  weight  of 

P  that  the  tension  may  be  1  pound  ;  5  pounds  ;  10 
pounds ;  20  pounds. 

6.  In  Fig.  21,  if  TF  =  20  pounds,  P  =  2  pounds ;    re- 

quired the  time  necessary  for  the  bodies  to  move 
10ft. 


[90.]  MEASURE    OF    FORCE.  51 


EXERCISES. 

1.  A  spring  balance  may  be  inserted  in  the  string  of  the  preceding 

problems  in  such  a  way  as  to  indicate  the  tension.  Suppose  that 
such  a  balance  were  inserted  in  the  vertical  part  of  the  string  in 
Fig:.  21,  and  the  string  cut  off  above  it,  what  will  the  balance  indi- 
cate afterwards  ? 

2.  If  a  man  pulls  a  weight  vertically  upward  by  means  of  a  cord 

attached  to  the  body,  will  the  tension  on  the  cord  equal  or  exceed 
the  weight  of  the  body  ; — the  weight  of  the  cord  being  neglected. 
Consider  the  case  when  the  motion  is  accelerated,  and  when  it  is 
uniform. 

3.  If  a  mau  stands  on  the  platform  of  an  elevator  as  it  descends  a  shaft 

with  an  acceleration,  how  will  his  pressure  upon  the  floor  compare 
with  his  weight  ?  How  if  the  platform  is  ascending  ? 

4.  If  a  man  slides  down  a  vertical  rope,  checking  his  velocity  by  grasp- 

ing it  more  or  less  firmly,  but  with  a  constant  grip,  how  will  the 
tension  on  the  rope  compare  with  his  weight  ? 


CHAPTER  III. 

WOKK FRICTION. 

91.  Work  is  the  overcoming  of  resistance  continually 
recurring  along  the  path  of  motion. — Thus  a  horse,  while 
drawing  a  load  on  a  cart,  does  work  by  constantly  over- 
coming the  friction  of  the  axle  and  the  resistances  of  the 
roadway.     The  same  effort,  however,  may  be  exerted  in 
producing  motion   only,  producing  live  or  stored  work. 
Hence,  the  following  is  a  more  general  definition  :  Work 
is  the  effect  produced  by  a  force  in  moving  its  own  point 
of  application  in  such  a  way  that  it  has  a  component 
motion  in  the  direction  of  action  of  the  force. 

92.  Measure  of  Work. — A  horse  that  draws  a  load 
two  miles  does  twice  the  work  of  drawing  it  one  mile,  and 
one-fifth  the  wrork  of  drawing  it  ten  miles.      The  work, 
therefore,   varies   directly  as  the   space  over  which  the 
resistance  is  overcome.    It  is  also  evident  that,  if  the  horse 
had  drawn  a  load  twice  as  large,  he  would  have  done  twice 
the  work  in  the  same  distance ;  hence,  the  work  also  varies 
directly  as  the  resistance.     This  principle  is  general,  and 
applies  to  all  cases  in  which  the  force  is  constant. 

Hence,  if 

F  =  a  constant  force  ; 
8  =  the  space  over  which  F  acts,  and 
U=  the  work  done  by  F\ 
then 

U=Fs. 

93.  The  work  of  a  variable  force  is  found  by  divid- 
ing the  space  into  small  parts,  so  small  that  the  force 


[94-96.]  WORK  53 

may  be  considered  constant  over  each  part,  and  taking 
the  sum  of  all  the  elementary  works. 
Let 

F  =  the  force  acting  over  any  one  of  the 

elementary  spaces; 
As  =  an  elementary  space  ; 
then 

{?= 


94.  Mere  motion  is  not  work.  —  If  the  planets  move  in 
space  without  meeting  any  resistance,  they  do  no  work. 

95.  The  Unit  of  Work  is  the  raising  of  one  pound  of 
matter  vertically  one  foot,  and  is  called  a  foot-pound. 
The  resistance  overcome  by  raising  matter  is  the  force  of 
gravity.     A  weight  of  one  pound  drawn  horizontally  is 
not  the  unit,  unless  the  frictional  resistance  should  happen 
to  equal  the  weight.     It  is  not  the  weight  moved,  but  the 
resistance  overcome,  that  constitutes  work. 

96.  Work  represented  by  a  Diagram.  —  When  the 
force  is  constant  the  work   may  be   represented  by  a 
rectangle. 

JLf 


••      abode 

PIG.  24.  FIG.  25. 

Thus,  in  Fig.  24,  let  the  base  of  the  rectangle  represent 
the  space  «,  and  the  altitude  represent  the  force  F,  then 
will  the  area  of  the  rectangle  be 

Fs, 
which  is  the  expression  given  in  Article  92  for  the  work. 

If  the  force  does  not  follow  a  known  law,  we  may  still 
find  the  work  approximately  by  constructing  a  curve,  the 
abscissas  of  which,  Aa,  ab,  be,  etc.,  shall  represent  the 


54  KINETICS.  [97,99.] 

spaces,  and  the  corresponding  ordinates,  am,  In,  co,  etc., 
shall  represent  the  corresponding  resistances ;  then  the  area 
of  the  figure  will  represent  the  work,  for  the  area  will  be 
*%FAs.     The  area  may  be  found  to  any  degree  of  approxi' 
mation  by  dividing  it  into  an  indefinite  number  of  trape- 
zoids,  finding  the  area  of  each  and  taking  their  sum. 
If  the  force  varies  directly  as  the  space  over  which  it 
c       acts,  the  work  may  be  represented 
by  the  area  of  a  triangle,  of  which 
the  base  represents  the  space  and 
the  altitude,  the  final  force.     For, 
any  ordinate  be  will  be  directly  as 


its  distance  from  A. 

97.  The  total  work  is  independ- 

ent of  the  time  required  to  perform  it.  —  For,  the  space 
may  be  passed  over  in  a  longer  or  shorter  time  without 
affecting  the  product  of  F  into  s.  The  horse  that  draws 
a  load  one  mile,  does  a  definite  amount  of  work,  whether 
it  be  done  in  one  hour  or  five  hours. 

98.  Time  and  velocity  are  implicitly  implied  in  doing 
work  ;  for,  the  space  involves  both.  If  the  space  be 
passed  over  with  a  constant  velocity  we  have 

*  =  vt, 
in  which 

v  =  the  velocity,  and 

t  =  the  time. 
Hence, 


therefore,  if  the  velocity  be  uniform,  the  work  will  vary 
directly  as  the  time  ;  and  if  the  time  be  constant,  the  work 
will  vary  directly  as  the  velocity. 
99.  Dynamic  Effect.  —  If  the  velocity  be  uniform  and 

t  =  1,  we  have  7        ^ 

work  =  l>v. 


[1UO-102.]  WORK.  55 

This  is  the  work  done  in  a  unit  of  time,  and  is  called 
Mechanical  Power,  or  Dynamic  Effect  and  sometimes 
simply  Power.  It  is  the  rate  of  doing  work,  or  simply 
work-rate. 

100.  The  unit  of  Dynamic  Effect  is  called  the  horse- 
power.    It  is  33,000  pounds  raised  one  foot  per  minute. 
It  was  determined  by  Boulton  &  Watt  by  means  of  trials 
with  horses  at  a  colliery  in  England.     It  is  doubtless  much 
larger  than  what  the  average  of  good  horses  can  do  for 
hours  in  succession,  but  it  may  be  considered  as  an  arbi- 
trary unit,  by  which  the  work  done  by  a  horse  or  other 
working  power  in  a  given  time,  may  be  measured.     It  is 
used  to  measure  the  efficiency  of  hydraulic  motors,  steam 
and  air  engines  and  other  machinery. 

101.  Work   may  be  useful  or  prejudicial. — That   is 
useful  which  produces  the  article  sought,  and  that  which 
wears  out   the  machinery   is  prejudicial.      The  former 
produces  money  for  the  mechanic  while  the  latter  costs 
money. 

Prejudicial  work  is  generally  f fictional  in  its  character, 
but  all  friction  is  not  prejudicial.  Thus,  the  friction  be- 
tween the  driving  wheels  of  a  locomotive  and  the  track  is 
necessary,  and  hence,  useful.  In  a  similar  way  the  fric 
tion  between  belts  and  pulleys  is  useful. 

It  is  not  always  possible  to 
distinguish  between  useful  and 
prejudicial  work.  The  latter 
always  accompanies  the  former, 
but  we  know  that,  for  economy, 
the  latter  should  be  reduced  as  Fia  ^ 

much  as  possible. 

102.  If  the  force  acts  at  an  angle  with  the  line  of 
motion  it  may  be  resolved  into  two  forces,  one  of  whicb 


56  KINETICS.  [103,104.] 

is  parallel  to  the  path  described  by  the  body  and  the  other 
perpendicular  to  it.  Thus,  in  Fig.  27,  the  normal  com- 
ponent of  the  force  F  is 


and  the  horizontal  component  is 

P  =  Fcos  FOP. 

There  being  no  motion  normal  to  the  path,  the  former 
component  does  no  work  ;  and  the  work  done  by  the  latter, 
according  to  the  definition,  will  be 

U-  Ps  =  Fsco&FOP. 

If  a  body  is  moved  around  a  semi-circumference  by  a 
constant  force  acting  parallel  to  a  fixed  diameter,  the 
work  will  be  the  product  of  the  force  into  the  diameter. 

103.  Work  in  a  Moving  Body.  —  A  moving  body  can- 
not be  instantly  brought  to  rest,  and  when  both,  resistance 
and  space  are  involved  in  a  result,  work  has  been  done. 
The  work  which  a  moving  body  is  capable  of  doing  equals 
the  product  of  the  mean  resistance  which  it  overcomes 
into  the  space  over  which  it  works.     Thus,  if  a  cannon 
ball  should  penetrate  the  earth  10  feet,  and  the  mean 
resistance  were  500  pounds,  the  work  done  would  be  5,000 
foot-pounds.     Another  mode  of  measuring  it  is  given  in 

Article  111. 

o 

1  ^  Friction. 

104.  Friction  is  that  force  or  resistance  between  two 
bodies  which  prevents,  or  tends  to  prevent,  one  body  from 
being  drawn  upon  another. 

All  bodies  are  rough.  However  perfectly  they  may  be 
polished,  an  inspection  of  their  surfaces  with  a  glass  of 
high  magnifying  power,  shows  that  they  are  still  very 
rough.  A  smooth  surface  is  a  comparative  term,  implying 


[105,  106.  ]  FRICTION.  .57 

that  it  is  more  or  less  smooth.  A  perfectly  smooth  surface 
probably  does  not  exist ;  but  when  the  term  is  used  it 
means  that  the  surface  offers  no  resistance  of  any  kind 
It  is  an  ideal  surface. 

It  is  certain  that  if  two  perfectly  smooth  plane  surfaces 
were  brought  in  contact,  they  would  offer  a  great  resist- 
ance to  being  drawn  upon  one  another  on  account  of  the 
adhesion  between  the  bodies.  They  would  hold  to  one 
another  nearly  as  strongly  as  if  they  were  solid.  When 
we  refer  to  smooth  surfaces  in  problems,  this  force  is  also 
excluded. 

105.  Experiments  in  regard  to  Friction. — M.  Moriu, 
a  French  savant,  was  one  of  the  first  to  determine  the  laws 
of  friction.     These  laws  were  deduced  by  experiments 
upon  a  variety  of  substances  under  a  variety  of  conditions. 
A  device  similar  to  that  shown  in  Fig.  29  was  used.     A 
body  W  was  placed  upon  a  long  strip  of  another  body,  and 
it  was  desired  to  determine  the  friction    between   them. 
A  strong  fine  cord  was  attached  to  the  body,  and,  passing 
over  a  pulley  at  the  end  of  the  platform,  was  attached  to 
a  dish  in  which  were  placed  weights  P.     The  weights  P 
were  made  to  exceed  slightly  the  frictional  resistance,  and 
thus  pull  the  body  W  along  the  other  body.     The  space 
over  which  they  moved  in  a  given  time  was  then  observed, 
and  with  the  data  thus  obtained  the  friction  was  computed 
as  shown  in  Article  109. 

Weights  were  added  to  the  body  W  so  as  to  produce 
greater  pressure  ;  also  weights  were  added  to  the  dish  and 
taken  from  it,  so  as  to  produce  a  greater  or  less  velocity. 

Different  substances  were  used  for  the  body  W,  and 
also  for  the  horizontal  strip  on  which  the  body  W  slid. 

106.  Angle  of  Friction. — Suppose  that  a  body  W  is 
placed  on  an  inclined  plane  A  C,  one  end  of  which  i? 

" 


58  KINETICS.  [107.] 

gradually  raised  until  motion  begins,  or  the  body  is  in  a 
state  bordering  on  motion.  The  weight  TFmay  be  resolved 
into  components  ;  one,  parallel  to 
the  plane,  which  tends  to  pull  the 
body  down  it,  and  the  other  per- 
pendicular to  it. 

Draw  Oa  parallel  to  AC,  and 
ab  perpendicular  to  it;   and  let 


FlCK  as.  Ob  represent  W.    Then,  according 

to  Article  52,  we  have 

F=Oa  =  TTsin  abO  =  Fsin  CAB. 

This  equals  the  resistance  due  to  friction. 

Let 

JV=  the  normal  pressure  =  ab  ; 
then 

jy=  TFcos  abO  -  IFcos  CAB  ; 
hence, 

F        TFsin  CAB 


which  is  called  the  angle  of  friction,  or  angle  of  repose. 

107.  Laws  of  Friction.  —  The  following  laws  have  been 
deduced  from  the  experiments  of  Morin  and  others  : 

1.  Friction  of  motion  is  slightly  less  than  that  of  rest. 

2.  The  total  amount  of  friction  is  independent  of  the 
extent  of  the  surfaces  in  contact. 

3.  The  amount  of  friction  between  two  surfaces  varies 
directly  as  the  normal  pressure,  and  with  the  character  of 
the  surfaces  in  contact. 

4.  Sliding  friction  is  independent  of  the  velocity. 
These  laws  are  sufficiently  accurate  for  ordinary  veloci- 

ties when  the  surfaces  do  not  abrade,  or  cut  one  another. 
Lubricants  aro   employed   to   diminish  friction.      Oil    if 


[108,109.]  FRICTION.  (      59 

the  most  common  lubricant,  though  water  is  better   in 
some  cases. 

108.  Coefficient  of  Friction. — According  to  the  third 
principle  of  the  preceding  article,  it  follows  that  the  ratw 
of  the  total  friction  to  the  total  normal  pressure  between 
two  surfaces  is  constant.     This  ratio  is  called  the  Coeffi- 
cient of  Friction. 

Let  N  =  the  normal  pressure ; 

F=  that  force  which  is  just  sufficient  to 
produce  motion  when  acting  par- 
allel to  the  plane  of  the  surfaces ; 
fi  =  the  Coefficient  of  friction  ; 
then 

F 

»  =  w-- 

Comparing  this  with  the  equation  of  Article  106,  we 
Bee  that  the  coefficient  of  friction  equate  the  tangent  of 
the  angle  of  repose. 

If  the  body  moves  on  a  horizontal  plane,  the  normal 
pressure  equals  the  weight ,  hence, 

F 

t  W' 

If  W  =  1,  /u-  =  F\  hence,  the  coefficient  of  friction 
equals  the  friction  caused  by  one  pound  of  the  body. 

109.  To  find  the  value  of  the  coefficient  of  friction 
of  motion. 

Let    W  =  the  weight  of  the  body  on  the  plane, 
P  —  the  weight  which  moves  it, 
y=  the  acceleration, 
/j,  =  the  coefficient  of  friction ; 
then  the  total  friction  will  be 


60  KINETICS.  [109.] 

and  the  effective  moving  force,  neglecting  the  motion  of 
the  pulley  and  the  cordage,  will  be 


and  the  mass  moved  will 
be 


Fio.  29. 


ff^\ 

P 


hence,  according  to  Arti- 
cle 86, 


'      '      P  +  W     ' 
This  value  in  equation  (4),  of  Article  24,  gives 


P+  W       > 
from  which  we  find 


Knowing  the  weights  P  and  W,  and  measuring  care 
fully  the  space  s  and  time  t,  the  second  member  becomes 
completely  known  ;  and  hence,  by  reduction,  fi  becomes 
known. 

For  oak  on  oak,  /*  =  0.48  when  the  fibres  are  parallel 

to  the  motion,  and 
0.19  when   the  fibres  are  perpen- 
dicular to  the  motion. 

"Wrought  iron  on  cast  iron,  /j,  =  0.18. 

Cast  iron  on  cast  iron,  p,  =  0.15. 

M.  Pambour  made  experiments  upon  the  frictional  re- 
sistance of  trains  of  cars  on  some  railroads  in  England,  and 


[110.]  FRICTION.  61 

found  the  friction  to  be  8  pounds  per  ton  gross  ;  hence, 
the  coefficient  was 

8  1 


__ 

~  2240  ~  280  * 

Experiments  in  this  country  gave  6£  pounds  per  ton 
gross  under  favorable  conditions. 

110.  Problems.  —  1.  A  piston  is  moved  in  a  horizontal 
cylinder,  as  shown  in  Problem  1,  Article  89,  by  a  constant 
steam  pressure  of  F  pounds.  At  what  point  must  the 
pressure  be  instantly  reversed  so  that  the  full  length  of 
the  stroke  shall  be  a  inches,  there  being  a  constant  friction 
of  F±  pounds  throughout  the  stroke? 

The  effective  driving  force  will  be  F  —  F±  pounds. 

The  effective  stopping  force  will  be  F  +  Impounds. 

Let 

*  =  the  space  over  which  the  former  acts,  and 
*!  =  the  space  over  which  the  latter  acts  ; 
then 

s  +  Si  =  a. 

The  work  done  upon  the  piston  by  the  driving  force 
will  equal  that  done  by  the  stopping  force  ;  hence, 


Eliminating  ^  between  these  equations,  gives 


2.  A  stream  of  waterfalls  over  a  dam  hfeet  high,  and 
has  a  section  of  a  square  feet  at  the  foot  of  the  fall  / 
required  the  horse-power  constantly  developed. 

(The  section  is  taken  at  the  foot  of  the  fall,  for  the 
velocity  is  measured  at  that  point.) 


62  KINETICS. 

The  velocity  at  the  foot  of  the  fall  will  be 


and  the  volume  of  water  which  passes  over  the  dam  in  one 
seer  nd  will  be 


The  weight  of  a  cubic  foot  of  water  being  62^  pounds, 
the  weight  of  the  quantity  will  be 

W=  62%a  i/fyh  pounds  per  second, 

=  60  x  62£«  \/%qh  pounds  per  minute. 

The  weight,  multiplied  by  the  height  A,  through  which 
it  falls,  will  give  the  work  it  can  do  in  one  minute,  and 
the  result,  divided  by  33,000,  will  give  the  horse-power  ; 


3.  Find  the  work  necessary  to  draw  a  body  up  an  in 
dined  plane. 

In  Fig.  28,  Article  106,  let 

l  —  AC\  b  =  AB  ;  h  =  BG,  W=  the  weight  of  the  body. 
As  shown  in  Article  106,  the  normal  pressure  will  be 


=      W, 

L 

and  hence,  the  friction  will  be 


which,  multiplied  by  the  length  of  the  plane,  gives  the 
work  necessary  to  overcome  the  friction  ; 


[110.]  FRICTION.  63 

The  component  of  the  weight  along  the  plane  will  be 

TFsin 


and  the  work  of  overcoming  the  weight  will  be  I  timee 
this  result,  or 


hence,  the  total  work  will  be 


hence,  it  equals  the  work  which  wotild  be  necessary  to 
draw  the  body  horizontally  from  A  to  J2,  and  lift  it  verti- 
cally from  B  to  C. 

4.  Required  the  work  necessary  to  compress  a  coiled 
spring  a  given  amount. 

It  is  found  by  experiment  that,  as  long  as  the  elasticity 
of  a  spring  remains  perfect,  the  amount  of  compression 
varies  directly  as  the  compressing  force.  That  is,  if  one 
pound  compresses  it  one  inch,  two  pounds  will  compress 
it  two  inches,  and  so  on.  Hence,  if 

p  =  the  force  which   compresses  a  spring 

one  inch, 

P  =  the  total  compressing  force, 
s  =  the  amount  of  the  compression  produced 

byP; 
then 


and  the  work  is  represented  by  Fig.  26,  in  which  AB  —  *, 
=  P. 


64  KINETICS.  [HO.1 

EXAMPLES. 

1.  How  many  cubic  feet  of  water  will  a  50  horse-power 

engine  raise  in  an  hour  from  a  mine  500  feet  deep. 
if  a  cubic  foot  of  water  weighs  62^  pounds  ? 

2.  Find  the  work  necessary  to  raise  the  material  in  making 

a  well  20  feet  deep  and  3  feet  in  diameter,  if  the 
material  weighs  140  pounds  per  cubic  foot. 

3.  The  pressure  on  a  steam  piston,  moving  horizontally,  as 

in  Prob.  1,  Art.  89,  is  1,000  Ibs.,  the  friction  200  Ibs. ; 
how  far  must  the  pressure  act  before  it  is  instantly 
reversed  that  the  full  stroke  may  be  12  inches  ? 

4.  The  French   unit  of  work  is  one  kilogramme  raised 

vertically  one  metre ;  required  the  equivalent  in 
foot-pounds.  (Take  the  metre  at  39.37  inches  and 
the  kilogramme  at  2.2  pounds.) 

5.  According  to  Navier  it  requires  43,333  French  units  of 

work  to  saw  a  square  metre  of  green  oak;  how 
many  foot-pounds  will  be  required  to  saw  a  square 
foot  of  the  same  material  ? 

6.  A  stream  of  water  falls  vertically  over  a  dam  12  feet 

high,  and  has  a  transverse  section  of  one  square  foot 
at  the  foot  of  the  fall ;  required  the  horse-power 
constantly  developed. 

7.  A  hammer,  whose  weight  is  2,000  pounds,  falls  verti- 

cally 8  feet;  how  far  will  it  drive  a  pile  into  the 
earth  if  the  constant  resistance  is  10,000  pounds  ? 

8.  If  it  is  found  by  means  of  a  spring  balance  that  a  span 

of  horses  pull  with  a  constant  force  of  200  pounds 
in  drawing  a  plough ;  if  they  travel  at  the  rate  of 
2  miles  per  hour,  what  will  be  the  mechanical  power 
required  to  work  the  plough  ? 

9.  In  Fig.  29,  let  W  =  40  pounds,  P  =  8  pounds,  and  ii 


[110.J  EXERCISES.  65 

is  observed  that  P  moves  over  4  feet  in  3  seconds ; 
required  the  coefficient  of  friction. 

10.  In  Fig.  29,  W  —  25  pounds,  P  —  5  pounds,  and  the 
coefficient  of  friction  =  0.15 ;  required  the  space 
over  which  the  bodies  will  pass  in  5  seconds. 


EXERCISES. 

1.  What  is  the  unit  of  work  ?    Is  work  the  same  as  force  ? 

2.  Is  friction  force  ?    When  is  it  useful  and  when  prejudicial  ? 

3.  In  what  sense  is  work  independent  of  the  time,  and  under  what  cir- 

cumstances is  it  dependent  upon  the  time  ? 

4.  A  body  placed  on  a  plane,  which  is  elevated  at  an  angle  of  15  degrees, 

is  just  on  the  point  of  moving ;  required  the  coefficient  of  friction 

Between  the  body  and  the  plane. 
6.  A  body,  whose  weight  is  25  pounds,  is  on  a  horizontal  plane ;  re 

quired  the  tension  of  a  string  by  which  the  body  is  drawn  alonj 

uniformly,  the  coefficient  of  friction  being  &. 
6.  Define  mechanical  power.     Is  it  work  ? 


X  CHAPTER  IV. 

ENERGY. 

111.  Energy  is  a  term  to  express  the  ability  of  an  agent 
to  do  work.     We  have  seen,  Article  103,  that  a  moving 
body  is  capable  of  doing  work.     A  slight  consideration  of 
bodies  at  rest  shows  that  they  are  also  capable  of  doing 
work.     Thus,  the  water  in  a  mill-pond  is  capable  of  doing 
useful  work  by  being  passed  through  a  water-wheel.     To 
do  work  the  water  must  be  in  motion,  but  the  weight  of 
the  water  falling  through  a  given  height  will  do  a  certain 
amount  of  work,  and  this  amount  can   be   determined 
while  the  water  is  in  position.     Similarly,  the  same  can 
be  shown  in  regard  to  other  bodies  at  rest.     These  ideas 
have  given  rise  to  the  terms  Kinetic  energy  and  Potential 
energy. 

112.  Kinetic  Energy  is  the  energy  of  a  moving  ~body, 
and  is  the  work  which  the  body  must  do  in  being  brought 
to  rest.     It  is  visible  energy. 

The  work  which  a  moving  body  is  capable  of  doing  is 
the  same  as  if  it  had  fallen  in  a  vacuum  through  a  height 
sufficient  to  produce  the  same  velocity. 
Let 

W  =  the  weight  of  a  body, 
v  —  its  velocity, 

h  =  the  height  through  which  the  body  must  fall 
to  produce  the  velocity  v. 


[118.1.  ENERGY.  67 

The  work  necessary  to  raise  a  body  a  height  h  will  be, 
according  to  Article  92, 

W/i. 

If  the  body  fall  freely  through  the  same  height,  it  will 
be  capable  of  doing  the  same  amount  of  work  when  it 
reaches  the  foot  of  the  fall.  The  velocity  produced  in 
falling  a  height  h  will  be  (Article  72), 


Substitute  this  value  in  the  preceding  expression,  and 
making, 


W 

M= —  (Article  80), 


we  have, 


v2  W 

Work  =  W/i  =  W^-  =  %—vi  =  %Mv*  =  K. 


The  expression  %Mv2  is  called  the  kinetic  energy,  and  is 
represented  by  J5T,  the  initial  letter  of  kinetic.  It  is  also 
called  the  vis  viva*  or  living  force  of  the  body.  Hence, 
the  kinetic  energy  of  a  moving  body  equals  the  work 
stored  in  it. 

113.  Potential  Energy  is  latent  energy.  It  is  the  work 
whicli  a  body  is  capable  of  doing  in  passing  from  one 
condition  or  position  to  another.  Thus,  the  power  in  a 
coiled  spring  is  potential;  but,  when  freed  from  its  restrain- 
ing power,  it  may  move  the  wheels  of  a  watch,  or  clock,  or 
drive  other  machinery,  in  doing  which  it  is  changed  from 
a  condition  of  tension  towards  one  free  from  tension.  The 

*  Mv-  is  often  called  the  vis  viva,  but  its  use  is  not  quite  as  convenient 
as  the  definition  in  the  text,  and  there  is  a  growing  tendency  towards 
the  general  adoption  of  the  definition  given  above.  It  makes  no  differ- 
ence, however,  which  is  used,  provided  it  is  always  used  in  the  same  sense. 


68  KINETICS.  [114.] 

power  in  a  weight  held  at  a  given  height  is  potential;  but, 
in  descending,  it  may  be  made  to  turn  machinery  and  thus 
do  work,  in  doing  which  the  body  passes  from  a  high  posi- 
tion to  a  lower  one.  The  power  stored  in  coal,  wood,  or 
other  fuel  is  potential ;  but,  if  the  fuel  be  burned,  it  may 
generate  steam  and  thus  do  work,  in  doing  which  it  i? 
changed  from  the  condition  of  fuel  to  that  of  ashes,  cinders, 
smoke,  etc.  The  power  in  gunpowder  is  l-atent  /  but,  if  the 
powder  be  exploded,  it  will  do  work,  and  may  be  made  to 
throw  a  cannon  ball,  or  rend  rocks,  or  produce  motive 
power.  The  power  contained  in  food  is  potential;  but 
the  food,  by  nourishing  animals,  and  thus  imparting 
strength  to  them,  becomes  a  source  of  work.  The  power 
contained  in  zinc  is  potential ;  but  the  zinc,  when  acted 
upon  by  acids,  becomes  active  and  capable  of  driving 
electro-magnetic  engines.  Air  compressed  and  stored  in 
a  vessel  contains  potential  energy  ;  but,  by  acting  upon 
suitable  machinery  as  it  expands  itself,  it  will  do  work. 
The  power  of  steam  confined  in  a  boiler  is  potential ;  but, 
if  the  steam  be  passed  through  suitable  mechanism,  it  will 
do  work. 

The  expression,  Change  of  Position,  is  sufficiently 
comprehensive  to  express  all  the  changes  of  condition. 
Potential  Energy  has  been  defined  as  Energy  of  Position. 
It  is  represented  by  the  Greek  letter  JI,  the  initial  of 
potential. 

Heat  Energy. 

114.  General  Statement. — It  is  well  known  that  heat 
may  be  produced  by  friction,  or  rather,  according  to  onr 
present  knowledge  of  the  subject,  the  work  of  overcoming 
friction  produces  heat.  Thus,  the  boy,  by  rubbing  a 
brass  button  briskly  on  his  sleeve,  soon  makes  it  too  hot 


[115,116.]  HEAT  ENERGY.  69 

for  the  comfort  of  his  neighbor's  hand.  Rubbing  two 
sticks  together  makes  them  warm.  One  of  the  earliest 
methods  of  obtaining  heat  was  by  friction.  The  heat 
produced  by  the  friction  of  a  match,  when  rubbed  on  a 
rough  surface,  ignites  the  phosphorus,  which,  by  burning, 
BO  increases  the  heat  as  to  set  the  wood  on  fire.  Axle 
bearings  in  machinery  often  become  so  hot  from  friction 
as  to  set  fire  to  the  oil  and  wood  which  surround  them. 
In  an  experiment  made  by  Sir  Humphrey  Davy  in  1799, 
two  pieces  of  ice,  rubbed  together  in  vacuo  at  a  tempera- 
ture below  32°  F.,  were  melted  by  the  heat  developed  at 
the  surfaces  of  contact. 

Iron  and  other  substances  may  be  heated  by  being 
struck  rapidly.  Compressing  air,  or  other  gaseous  bodies, 
develops  heat. 

115.  The   Dynamic  theory  of  heat  rests  upon   the 
hypothesis  that  heat  consists  of  the  motion  of  the  mole- 
cules of  a  body,  or  is  the  result  of  that  motion,  and  that  to 
produce    these    motions    requires  a   definite   amount  of 
mechanical  energy.     This  hypothesis  is  confirmed  by  the 
experiments  of  Joule,  who  produced  heat  in  a  variety  of 
ways — compressing  air,  compressing  gases,  agitating  water, 
and  by  friction  between  cast-iron  surfaces. 

According  to  Joule's  experiments,  if  a  body  weighing 
one  pound  were  permitted  to  fall  freely  772  feet  in  a 
vacuum,  and  all  the  energy  thus  acquired  could  be  utilized 
in  heating  one  pound  of  water,  it  would  raise  the  tempera- 
ture 1°  Fahrenheit.  This  is  Joule's  equivalent,  and  in  the 
mathematical  theory  of  heat  is  represented  by  J.  It  is 
recognized  as  the  mechanical  equivalent  of  heat. 

116.  The  Mechanical  Equivalent  of  Heat. — There  ia 
a  definite  relation  between  the  work  expended  in  over- 
coming friction  and  the  heat  which  is  produced  by  it. 


TO  KINETICS.  [116.1 

The  earliest  experiments  of  which  we  have  any  historical 
knowledge  tending  to  prove  this  fact  were  made  by  Count 
Rumford.  In  1798  he  published  in  the  Trans,  of  the  Royal 
Phil.  Society  some  of  his  experiments  upon  this  subject. 
A  brass  cannon  weighing  113  pounds  was  revolved  hori- 
zontally, at  the  rate  of  32  revolutions  per  minute,  against 
a  blunt  steel  borer  with  a  pressure  of  10,000  pounds.  In 
half  an  hour  the  temperature  of  the  metal  had  risen  from 
60°  to  130°  F.  This  heat  would  have  been  sufficient  to 
raise  the  temperature  of  five  pounds  of  water  from  32°  to 
212°.  In  another  experiment  the  cannon  was  placed  in  a 
vessel  of  water  and  friction  applied  as  before.  In  two 
hours  and  a  half  the  water  actually  boiled.  The  heat 
generated  in  this  case  was  calculated  by  Rumford  to  be  at 
least  equal  to  that  given  out,  during  the  same  time,  by  the 
burning  of  nine  wax  candles,  three-quarter  inch  in  dia- 
meter, each  weighing  245  grains. 

Fourrier,  in  the  year  1807,  gave  the  laws  of  the  trans- 
mission of  heat  by  radiation  and  conduction,  and  laid  the 
foundation  for  the  mathematical  theory  of  heat.  Sadi 
Carnot,  in  his  work  entitled  "  Reflexions  sur  la  puissance 
motrice  du  feu,"  published  in  1824,  compared  the  energy 
of  heat  to  that  of  a  fall  of  water  from  one  level  to  another. 

This  branch  of  science,  however,  made  little  or  no  prog- 
ress until  it  was  shown  that  there  was  a  definite  relation 
between  heat  and  work.  Several  contemporaneous  investi- 
gators entertained  the  view  that  such  definite  relation 
existed,  and  labored  independently  to  prove  it.  In  1842 
Dr.  Mayer,  a  physician  of  Heilbronn,  formally  stated  that 
there  exists  a  connection  between  heat  and  work,  and  first 
introduced  the  expression,  ^'mechanical  equivalent  of 
heat"  In  the  same  year  Colding,  of  Copenhagen,  pub- 
lished experiments  on  the  production  of  heat  by  friction, 


[117.] 


HEAT    ENERGY. 


71 


from  which  he  concluded  that  the  quantity  of  heat  pro- 
duced by  friction  was  directly  proportional  to  the  work 
expended. 

But,  the  most  important  were  the  labors  of  Dr.  J.  P 
Joule,  of  Manchester,  England,  who,  during  the  years  1840 
to  1843,  by  a  series  of  very  careful  and  elaborate  experi- 
ments, determined  a  value  for  the  mechanical  equivalent 
of  heat  which  is  considered  as  the  most  reliable  ever 
found.  (See  Phil.  Trans.,  1850,  p.  61.) 

117.  Joule's  Experiments. — It  being  impracticable  to 
change  the  energy  of  a  body  falling  freely  into  heat  in 
such  a  way  as  to  measure  the  exact  equivalent,  Joule 
resorted  to  different  devices.  One  of  the  most  reliable  is 
the  following : 

A  copper  vessel  J2,  Fig.  30,  was  filled  with  water  and  pro- 
vided with  a  brass  paddle-wheel  (shown  by  dotted  lines), 
which  could  be  made  to  rotate  about  a  vertical  axis.  The 


FIG. 


paddle  had  numerous  openings,  so  as  to  agitate  the  water 
as  much  as  possible  when  it  rotated  in  the  vessel.  The  vessel 
was  closed  so  as  to  prevent  the  escape  of  the  water,  and  was 


72  KINETICS.  [118.] 

provided  with  a  thermometer  to  measure  a  change  of  the 
temperature.  Two  weights,  E  and  2<]  were  attached  to 
cords  which  passed  over  the  axles  of  the  pulleys  C  and  D. 
and  were  connected  with  the  axis  A,  so  that  as  the  weighte 
descend  the  paddle  would  be  made  to  revolve.  The 
height  of  the  fall  was  indicated  by  the  scales  G  and  IT, 
the  total  fall  in  Joule's  experiments  being  about  63  feet. 
The  roller  A  was  so  connected  to  the  axis  of  the  paddle 
that,  by  removing  a  pin,  the  weights  could  be  wound  up 
without  disturbing  the  axle.  In  this  way  the  experiments 
were  repeated  twenty  times. 

The  work  done  by  gravity  was  expressed  by  (F+  E)xh. 
This  was  all  expended  in  the  following  ways:  1st,  chiefly 
in  overcoming  the  resistance  of  the  water ;  2d,  in  over- 
coming the  friction  of  the  several  bearings ;  and  lastly, 
the  kinetic  energy  in  the  moving  parts  at  the  instant 
the  motion  was  stopped.  In  the  experiments  the  two 
latter  were  reduced  as  much  as  possible  by  the  mechani- 
cal arrangements,  but  their  effects  were  computed  and 
deducted  from  the  work  done  by  gravity. 

The  remaining  work  was  expended  in  overcoming  the 
resistance  of  the  water,  and  thus  developed  heat. 

Taking  as  a  unit  of  heat  that  necessary  to  raise  the 
temperature  of  one  pound  of  water  1°  F.,  the  unit  of  heat 
equals  772  foot-pounds  of  work. 

If  the  unit  of  heat  be  that  necessary  to  raise  one  kilo- 
gramme of  water  1°  C.,  then  a  unit  of  heat  equals  424 
kilogramme-metres. 

General  Principles  of  Energy. 

118.  Transmutation  of  Energy. — We  have  seen  that 
kinetic  energy  may  be  changed  into  an  equivalent  of  heat 


[118.]  PRINCIPLES    OF    ENERGY.  73 

energy  through  the  agency  of  friction.  We  also  know 
that  heat  energy  may  be  changed  into  kinetic  energy,  as 
is  constantly  done  in  the  ordinary  steam-engine.  Other 
energies  are  constantly  brought  into  action,  such  as  elec- 
tricity, magnetism,  chemical  action,  forces  of  polarity,  and 
indeed  any  agency  by  which  matter  is  moved.  The  rela- 
tions between  the  different  kinds  of  energies  are  not  well 
known,  except  that  between  heat  and  visible  energy  ;  but 
it  is  believed  that  they  all  consist  of  some  kind  of  mole- 
cular motion,  and  are  kinetic  or  potential  according  to 
circumstances,  and  that  they  are  changeable  one  into  the 
other. 

This  transmutation  is  constantly  going  on.  To  illus- 
trate it  with  an  example :  The  energy  of  the  sun's  heat  is 
stored  in  the  plant  and  there  becomes  potential.  The 
plant  may  be  changed  to  coal,  imparting  some  of  its 
energy  to  surrounding  objects,  but  concentrating  its 
remaining  potential  energy  into  a  smaller  space.  Coal  is 
raised  from  its  bed  by  means  of  kinetic  energies,  and  used 
in  a  locomotive  engine,  for  instance,  where  it  is  burned 
and  becomes  kinetic.  One  portion  of  this  energy  becomes 
stored  in  the  steam  in  the  boiler ;  another  portion  escapes 
with  the  smoke  ;  still  another  portion  escapes  through  the 
walls  of  the  fire-box ;  and  another  is  thrown  away  with 
the  live  coals  which  fall  through  the  grate,  or  are  hauled 
out  of  the  door  of  the  fire-box. 

The  steam  from  the  boiler  is  admitted  into  the  cylinder 
of  the  engine,  and  a  portion  of  this  energy  is  utilized  in 
driving  the  piston,  and  another  portion  escapes  with  the 
exhaust  steam.  The  energy  of  the  piston  is  expended  in 
producing  heat  on  the  track,  heating  the  axles  under  the 
cars,  overcoming  the  resistance  of  the  air,  wearing  the 
couplings  between  the  cars  and  the  working  parts  of  the 


74:  KINETICS.  [119.  120J 

machinery,  bruising  the  ends  of  the  rails,  crushing  the 
ties  under  the  track,  disturbing  the  earth  or  other  material 
which  forms  the  roadway,  and  imparting  kinetic  energy  to 
the  train.  The  energies  which  are  transmitted  to  these 
several  elements  finally  disappear  in  heat,  which  is  quickly 
absorbed  by  the  atmosphere,  beyond  which  we  are  unable 
to  trace  it  with  any  degree  of  certainty. 

119.  Conservation  of  Energy. — These  principles  are 
included  in  a  general  law  called  the  Conservation  of 
Energy,  which  may  be  stated  as  follows  : 

1.  The  total  amount  of  energy  in  the  Universe  is  con- 
stant. 

2.  The  various  forms  of  energy  may  be  converted  the 
one  into  the  other. 

It  follows  from  the  former  of  these  that  no  energy  is 
ever  lost.  Energy  is  indestructible. 

This  law  is  the  result  of  a  long  series  of  observations, 
experiments,  and  generalizations,  but  it  is  now  considered 
as  firmly  established  as  any  law  in  nature.     It  is  accepted 
as  a  fundamental  law  in  Physical  Science,  and  is  as  uni- 
versal in  its  application  as  the  law  of  gravitation. 
!l    120.  Non-equilibrium   of  Energies. — Energies  work 
kraly  as  they  pass  from  one  condition  to  another,  and  this 
I' is  done  only  when  they  are  not  in  equilibrium.     Thus,  if 
/two  metals  are  equally  hot,  one  cannot  impart  heat  to 
|  the  other.     A  body  by  losing  heat  loses  energy.     When 
steain  works  by  expansion,  the  temperature  is  reduced. 
I  The  energy  stored  in  coal  is  developed  by  being  burned, 
•  but  the  heat  thus  produced  exceeds  that  which  it  imparts 
to  steain  or  to  other  bodies.     We  know  of  no  means  by 
which  heat  energy  can  be  changed  into  an  equivalent  of 
kinetic  energy  ;    for  in  every  attempt  to  accomplish   it 
there  is  an  apparent  loss  of  energy,  some  of  it  becoming 


[121.]  PRINCIPLES   OF   ENERGY.  75 

potential,  or  assuming  an  energy  of  a  lower  form.  This  is 
called  dispersion  of  energy.  We  have,  however,  seen  that 
visible  energy  is  readily  changed  into  heat.  Should  all 
the  energies  of  the  universe  finally  become  changed  into 
heat  uniformly  distributed  throughout  space,  all  motion 
would  cease,  and  the  universe  would  become  virtually 
dead.  Such  a  result  has  been  predicted  by  some  writers 
upon  this  subject. 

121.  Perpetual  Motion. — By  perpetual  motion,  in  a 
popular  sense,  is  not  meant  ceaseless  motion,  such  as  we 
see  in  the  earth  and  other  planets,  but  a  machine  which, 
when  put  in  motion,  will  continue  in  motion  indefinitely, 
without  the  application  of  additional  power.  But  every 
machine,  when  in  motion,  must  overcome  the  resistance  of 
the  air  and  the  friction  of  the  bearings,  and  as  these 
resistances  cannot  be  entirely  removed,  or  annihilated,  it 
necessarily  does  work,  thus  consuming  the  energy  imparted 
to  it.  When  all  the  energy  has  been  consumed  the  ma- 
chine will  stop  and  remain  at  rest.  A  perpetual  motion 
machine  is  an  impossibility.  In  order  to  be  possible  it 
must  expend  more  power  than  is  imparted  to  it ;  or  in 
other  words,  it  must  possess  a  self-creating  power. 

EXAMPLES. 

1.  How  many  foot-pounds  of  work  is  stored  in  a  body 

which  weighs  25  pounds,  and  has  a  velocity  of  J.OO 
feet  per  minute  ? 

2.  If  a  body,  moving  with  a  velocity  of  5  feet  per  second, 

penetrates  the  earth  2  feet,  how  far  would  the  same 
body  penetrate  it  moving  with  a  velocity  of  15  feet 
per  second,  the  resistance  of  the  earth  being  uniform 
along  the  path  of  the  body  in  both  cases  ? 


76  KINETICS.  [12'.-J 

3.  If  a  train  of  cars  weighs  60  tons,  and  moves  at  the  rate 

of  40  miles  per  hour,  how  far  will  it  move  before 
being  brought  to  rest  by  friction,  the  friction  being 
8  pounds  per  ton,  no  allowance  being  made  for  the 
resistance  of  the  air  \ 

4.  A  train  of  cars  weighing  200,000  Ibs.,  and  moving  at 

the  rate  of  20  miles  per  hour,  is  suddenly  stopped  ; 
if  all  its  energy  be  utilized  in  heating  water,  how 
many  pounds  of  water  would  be  raised  in  tempera- 
ture from  32°  F.  to  the  boiling  point,  or  to  212°  F.  ? 

5.  In  Fig.  29,  Article  109,  if    W=  10  Ibs.,  P  -  4  Ibs., 

fjt,  =.  0.2,  g  —  32£,  and  the  weight  W  is  drawn  5  feet 
by  the  weight  P  falling  the  same  distance,  when  the 
latter  strikes  the  ground ;  how  far  will  the  weight 
W  move  before  being  brought  to  rest  by  friction  ? 

6.  If  a  piece  of  iron  whose  weight  is  200  Ibs.  is  moved 

at  a  uniform  rate  to  and  fro  on  another  piece  of 
iron,  the  coefficient  of  friction  between  them  being 
0.2,  what  must  be  the  velocity  of  the  moving  body 
BO  that  the  heat  developed  by  the  friction  would,  if 
entirely  utilized,  raise  the  temperature  of  5  Ibs.  of 
water  50°  F.  every  3  minutes. 


EXEECISE8. 

1.  IB  force  the  same  as  energy  ? 

2.  Does  kinetic  energy  mean  work  done,  or  ability  to  do  work  ? 

3.  An  animal  eats  food  which,    mechanically  speaking,   is  potential 

energy  ;  indicate  some  of  the  energies  which  follow  the  digestion 
of  the  food. 

4.  Why  is  not  wood  heated  as  much  by  a  ball  fired  into  it  as  iron  is  bj 

a  ball  fired  against  it  ? 

5.  Will  a  ball  flying  through  the  air  be  wanned  on  account  of  the  Mo- 

tion of  the  air  ? 


[121.]  EXERCISES.  77 

6.  What  becomes  of  the  energy  due  to  the  motion  of  the  water  ID 

rivers  ? 

7.  Why  do  a  person's  hands  become  warm  by  rubbing  them  against  one 

another  ? 

8.  If  a  ton  of  coal,   which  costs  5  dollars,  will  evaporate  a  certain 

amount  of  water,  and  it  requires  two  cords  of  wood  to  evaporate 
the  same  amount,  what  will  the  wood  be  worth  per  cord  for  the 
same  purpose  ? 


CHAPTER  V. 

MOMENTUM. 

122.  The  equation  of  Article  86  is 


Equation  (1)  of  Article  24  is 
•  «>t{ 

and,  eliminatingyfrom  these  equations,  gives 
Ft  =  Mv. 

The  quantity  Mv  is  called  the  momentum  of  a  body 
whose  mass  is  M.  This,  however,  is  merely  giving  a 
name  to  an  expression,  but  by  comparing  it  with  the  first 
member  of  the  equation  we  see  that  it  is  the  effect  which 
a  constant  force  F  produces  in  a  time  t.  We  therefore 
call  it  a  time-effect,  and  represent  its  value  by  Q. 

.'.  Q  =  Mv.    .        .        .        (1) 

If  a  body  has  a  velocity  v0  when  the  force  begins  to  act, 
and  a  velocity  v  after  a  time  t,  then 

Q  =  Ft  =  M(v-v,).     .        .        (2) 

123.  Momentum  of  a  variable  force.  —  If  the  force 
be  variable,  suppose  that  the  time  is  divided  into  such 
email  portions  that  the  force  may  be  considered  as  con- 
stant during  each  portion,  then  will  the  total  effect  be  the 
sum  of  all  the  elementary  momenta. 


I  134-126.]  MOMENTUM.  79 

Let  di  =  an  element  of  time, 

F=  the  force  during  any  element  of  time  ; 
then 

Q  =  2F.  ~Ai  =  M(v  -  v0). 

In  these  equations  F  is  the  effective  moving  force. 

124.  Impulse.  —  An  impulse  is  the  time  effect  of  a  blow. 
When  one  body  strikes  another,  as  a  hammer  striking  an 
anvil,  its  effect  is  an  impulse.  The  body  struck  may  be 
fixed  or  free  to  move  ;  but  the  problems  which  are  usually 
considered  under  this  head  generally  pertain  to  those  in 
which  the  body  considered  is  free  to  move.  Although  the 
effect  will  be  produced  in  an  exceedingly  short  time,  yet 
the  result  is  a  time-effect,  and  is  measured  in  the  same  way 
as  an}'  other  time-effect.  In  the  case  of  a  blow,  the  pres- 
sure between  the  bodies  will  be  variable  during  contact  ; 
but  for  the  sake  of  making  a  practical  formula,  let 

F'=-  the  mean  pressure  between  the  bodies, 
M  =  the  mass  of  the  body  considered, 

v  =  the  velocity  produced  by  the  blow  ; 
then 

=  F't  =  Mv. 


125.  Momentum  is  not  a  force.  —  The  force  F  is  only 
one  of  the  elements  of  the  expression.     The  unit  of  mo- 
mentum is  one  pound  of  mass  moving  with  a  velocity  of 
one  foot  per  second. 

Momentum  is  sometimes  called  quantity  of  motion  and 
sometimes  quantity  of  velocity,  but  neither  fully  expresses 
its  meaning  The  expression  time-effect  is  partly  descrip- 
tive of  its  meaning,  and  is  preferred  to  either  of  the 
others. 

126.  Instantaneous  Force.  —  When  an    effect  is  pro- 
duced in  an  imperceptibly  short  time,  the  agency  which 


80  KINETICS.  [127,  128.J 

produces  it  is  sometimes  called  an  instantaneous  force,  a 
farm  which  implies  that  the  effect  is  produced  instantly, 
requiring  no  time  for  its  action.  No  force  prodiices  its 
effect  instantly,  and  hence  the  term  is  liable  to  mislead. 
Sometimes  it  is  called  an  impulsive  force,  but  this  is  also 
objectionable,  for  it  implies  that  the  effect  is  a  force^ 
whereas  the  effect  is  either  momentum  or  work.  On 
account  of  these  objections  it  appears  advisable  to  use  the 
term  impulse  instead  of  either  of  the  above. 

127.  Problems.  —  1.  If  a  body,  whose  weight  is  25  Ibs., 
is  drawn  along  a  horizontal  plane  by  a  constant  pull  of 
6  Ibs.,  the  coefficient  of  friction  being  y1^,  what  will  be  the 
velocity  at  the  end  of  t  seconds  f 

The  frictional  resistance  will  be 

rVof  25  Ibs.  =  2.5  Ibs.; 

hence,  the  effective  pulling  force  will  be 

F-  6  -  2.5  =  3.5  Ibs. 

Ft       3.5  x  193 
=  M~~    -v^-t----^tft'P«r*™- 

2.  Required  the  constant  force  necessary  to  impart  to  a 
body,  whose  weight  is  100  Ibs.,  a  velocity  v  during  5 
seconds. 

„     Mv         100 

"  y 


Impact. 

128.  Impact  is  the  impinging  of  one  body  against 
another. 

It  is  direct  when  the  line  of  motion  of  the  impinging 
body  is  normal  to  the  body  struck,  as  in  Figs.  31  and  32. 


[129.] 


ELASTICITY. 


81 


It  is  central  when  the  line  of  motion  passes  through  the 
Centre  of  the  body  struck,  as  in  Figs.  31  and  33. 

It  is  direct  and  central  when  the  line  of  motion  passes 
through  the  centre  of  the  body  struck,  and  is  normal  to 
the  surface  at  the  point  of  impact,  as  in  Fig.  31. 


Fio.  81. 


Flu, 


Pio.  38. 


It  is  oblique  when  the  line  of  motion  is  inclined  to  sur- 
face at  the  point  of  contact. 

It  is  eccentric  when  the  line  of  motion  is  normal  to  the 
surface  of  the  body  struck  at  the  point  of  contact,  but 
does  not  pass  through  the  centre  of  the  body,  as  in  Fig.  32. 


Elasticity. 

129.  Elasticity  is  that  property  of  bodies  by  which  they 
gain,  or  seek  to  gain,  their  original  form  after  they  have 
been  elongated,  compressed,  twisted,  bent,  or  distorted  in 
any  way.  In  order  to  discuss  problems  involving  impact, 
it  is  necessary  to  know  the  laws  of  elasticity.  It  is  found 
by  experiment  that,  if  a  body  be  pulled  by  a  force  in  the 
direction  of  its  length,  as  in  Fig.  34,  it  will  be  elongated  ; 
and  if  the  elongation  be  small  it  will  shorten  itself  when 
the  pulling  force  is  removed.  It  is  found  that  all  known 
substances  are  more  or  less  elastic.  Air  and  gases  are 
highly  elastic. 

If  bodies  regain  all  their  distortion  after  the  force  is 


82  KINETICS.  [130.] 

removed,  they  are  called  perfectly  elastic.  If  they  regain 
only  a  part  of  their  distortion,  they  are  called  imperfectly 
elastic,  and  if  they  regain  ncne  of  their  distortion,  they  are 
called  non-elastic.  The  perfect  elasticity  of  solids,  even 
within  small  limits,  has  been  questioned,  but  for  practical 
purposes  many  of  them,  such  as  steel,  good  iron,  glass, 
ivory,  etc.,  may  be  considered  as  perfectly  elastic.  Pris- 
matic bars  of  good  iron  and  steel  may  be  elongated  about 
•g-g^-g-  of  their  length  without  damaging  their  elasticity. 

130.  Coefficient  of  Elasticity.  —  Experiments  show 
that  the  elongation  or  compression  of  prismatic  bars  of  a 
solid,  within  small  limits,  varies  directly  as  the  pulling  or 
pushing  force,  and  inversely  as  the  transverse  section. 

The  coefficient  of  elasticity  is  the  pulling  force  per  unit 
of  sectijn  divided  by  the  elongation  per  unit  of  length. 

BC 


FIG.  34. 


If      I  =  AB  =  the  original  length  of  the  piece, 
\  —  J3C=  the  elongation, 
F=  the  pulling  force, 
7T=  the  transverse  section, 
E  =  the  coefficient  of  elasticity  ; 
then 

W 

-j~  =  the  strain  on  a  unit  of  section,  and 
ji. 

-5  =  the  elongation  per  unit  of  length  ; 

i 

and,  according  to  the  above  definition,  we  have 

F     \       Fl  w 

-=•  -v-  7  =  --  =  constant  =  A 

XL  i 


[131,132.]  ELASTICITY.  83 

Values  of  E  in  pounds  per  square  inch. 

For  wrought  iron  from  23,000,000  Ibs.  to  28,000,000  Ibs. 
For  steel  from  25,000,000  Ibs.  to  31,000,000  Ibs. 

For  wood  from  1,000,000  Ibs.  to   2,000,000  Ibs. 

131.  Elongation   of  a    prismatic    bar.  —  Solving   the 
equation  in  the  preceding  article,  \ve  have 


which  is  the  expression  sought.  This  expression  is  true 
only  for  such  strains  as  do  not  damage  the  elasticity. 

132.  Modulus  (or  Coefficient)  of  Restitution.—  If  an 
elastic  body  impinge  upon  another,  the  bodies  will  at  first 
compress  one  another,  the  compression  increasing  rapidly 
for  a  very  short  time  until  a  maximum  is  reached  ;  after 
which,  by  virtue  of  their  elasticities,  they  tend  to  regain 
their  original  form,  and  thus  force  themselves  apart.  The 
force  which  causes  them  to  separate  can  act  only  while 
they  are  in  contact  with  one  another,  but  they  may  con- 
tinue to  regain  their  form  after  separation. 

The  force  between  the  bodies  during  compression,  and 
also  during  restitution,  is  constantly  changing,  and  the  law 
of  change  may  be  very  complex.  The  effect  is  to  constantly, 
but  very  rapidly,  change  the  velocity  of  one  or  both  the 
bodies  during  contact.  If  the  bodies  are  composed  of  the 
same  substance,  and  the  impact  is  not  so  severe  as  to 
damage  their  elasticity,  the  velocity  gained  by  a  body  dur- 
ing restitution,  divided  by  the  velocity  lost  during  com- 
pression, will  be  constant,  and  is  called  the  modulus  of 
restitution,  or  index  of  elasticity.  It  is  represented  by  e. 
If  the  bodies  are  moving  in  the  same  direction,  the  body 
struck  will  gain  velocity  during  compression,  and  still 


84  KINETICS.  [132.1 

more  velocity  during  restitution.  In  this  case  the  modulus 
of  restitution  will  be  found  by  dividing  the  velocity 
gained  during  the  restitution  by  the  velocity  gained  during 
the  compression.  In  short,  it  is  the  ratio  of  the  effect  011 
the  velocity  due  to  restitution  to  that  due  to  compression. 
According  to  these  principles  a  simple  expression  may 
be  found  for  e  by  assuming  that  the  body  struck  is  at 
rest,  and  so  large  that  it  will  be  unaffected  by  a  blow  from 
the  moving  body  ;  for,  in  this  case,  the  velocity  at  the  in- 
stant of  greatest  compression  will  be  zero,  and  hence  the 
velocity  lost  during  compression  will  equal  the  velocity  of 
approach  of  the  striking  body,  and  the  velocity  regained 
will  be  that  with  which  it  rebounds. 

Let 

v  =  the  velocity  of  the  impinging  body  at  the  in- 
stant impact  begins,; 

v^  =  the  velocity  at  the  end  of  the  impact ; 
then, 

.  =  5    .      .  .    (1) 

V 

Let  the  moving  body  fall  upon  the  immovable  one,  and 

'let 

H=  the  height  of  the  fall, 
h  =  the  height  of  the  rebound ; 
then 


(2) 


In  this  way  the  values  of  e  have  been  found  for  the  fol- 
lowing substances  : 


[133.] 


IMPACT. 


86 


Substance. 

Value  of  «. 

Substance, 

Value  of  «. 

Glas8  

0.94 

Steel,  soft  

0.67 

Hard  baked  clay.  .  .  . 

0  89 

Cork  

0.65 

Ivory  

0  81 

0  41 

Limestone  

0  79 

0  20 

Steel,  hardened  
Cast-iron  

0.79 
0.73 

Clay,    just  yielding 
to  the  hand  

0.17 

For  perfectly  non- elastic  bodies  e  is  zero.  If  the  resti- 
tution were  perfect  during  contact  e  would  be  unity,  and 
the  bodies  would  be  called  perfectly  elastic.  The  momen- 
tum lost  by  a  body  during  compression  is  generally  called 
the  force  of  compression,  and  that  gained  during  restitu- 
tion the  force  of  restitution. 

133.  Problem. — Given  the  masses  and  velocities  of  two 
perfectly  free  non-elastic  bodies  before 
impact,  to  find  their  velocities  after 
impact. 

Take  the  simplest  case,  that  of  direct 
central  impact.  Let  the  body  Q  im- 
pinge upon  Qr,  the  motion  being  in 
the  same  direction. 

Let 

M=  the  mass  of  Q, 
v  =  the  velocity  of  Q  before  impact, 
M'  and  v'  the  corresponding  quantities  for  Q'}  and 
V  =  the  common  velocity  after  impact. 

The  momentum  lost  by  Q  during  impact  will  be 

Q  =  M(v-  F); 
and  that  gained  by  Q'  will  be 


FIG.  35. 


86  KINETICS.  [134-1 

Q'  =  M'(V-v'); 

which,  being  time-effects,  will  equal  one  another  ;  hence, 
M(v—  F)  =  M'(V-v') 
M'v' 


M  +  M'  '     '     ' 
Clearing  of  fractions  gives 

M'v'',    .     .    (2) 


that  is,  the  momentum  of  the  bodies  after  impact  equah 
the  sum  of  momenta  before  impact. 
To  find  the  velocity  in  terms  of  the  weights,  substitute 

W  W 

—  for  M  and  —  for  M  ',  in  equation  (1),  and  we  have 
y  y 

Wv  +  W'v' 
W+  W 

If  the  bodies  are  moving  toward  one  another  before 
impact  one  of  the  velocities  will  be  negative.  Making  v' 
negative,  we  have 

Mv-M'v'  . 
M  +  M'    '  ' 

in  which  if  Mv  ==  M'v',  we  have  V  =  0  ;  that  is,  if  two 
inelastic  bodies  of  equal  momenta  impinge  directly  upon 
one  another  from  opposite  directions  they  will  be  brought 
to  rest  by  the  impact. 

134.  Loss  of  velocity.  —  From  equation  (1)  of  the 
preceding  article  we  find,  by  subtracting  both  members 
from  -y, 


[135.]  IMPACT.  87 

and  similarly,  subtracting  both  members  from  vr  gives, 

«-T=-(?-^>  •  •  (2) 


which,  being  negative,  indicates  that  there  is  a  gain  ol 
velocity. 

135.  Impact  of  perfectly  elastic  bodies.  —  It  has 
been  observed  in  Article  132  that  at  the  instant  of  greatest 
compression  the  bodies  have  a  common  velocity  ;  hence, 
at  that  instant,  the  velocity  will  be  given  by  the  equations 
of  Article  133  ;  and  the  loss  of  velocity  up  to  that  instant 
will  be  given  by  the  equations  in  Article  134:.  Bat  if  the 
bodies  are  perfectly  elastic,  the  effect  upon  the  velocities 
during  restitution  will  be  exactly  the  same  as  during  com- 
pression •  hence,  the  final  loss  of  velocity  will  be  double 
that  during  compression  ;  therefore,  for  the  striking  body, 
it  will  be 

2JT      ,          - 

JT+JT(V-V^  ' 

and  for  the  body  struck  it  will  be 


These  subtracted  from  the  velocity  before  impact  will 
give  the  final  velocity. 
Let 

0X  =  the  velocity  of  the  former  body  after  impact, 

and 

v\  =  the  velocity  of  the  latter  after  impact  ; 
then, 


88  KINETICS.  [13G.J 


and  these  may  be  reduced  to 

Mv  +  M'v'  M' 


'(V- 


,     Mv  +  M'v'  M 

^  +  -'(v- 


wtlch,  by  means  of  equation  (1)  of  Article  133,  become 


which  show,  as  they  should,  that  the  final  velocity  equals 
the  common  velocity  at  the  instant  of  greatest  compres- 
sion, increased  or  diminished,  as  the  case  may  be,  by  the 
velocity  due  to  restitution. 

136.  Discussions  of  Equations  (3)  and  (4)  of  the  pre- 
ceding article. 

1st.  Let  M  =  M',  then  we  have 

Vi  =  v  —  (v  —  vf)  =  v'  ; 
Vi'=  v'+  (v  —  v')  =  v  ; 

that  is,  they  will  interchange  velocities. 
2d.  Let  v'  =  0  and  M  =  M't  then 


that  is,  the  first  body  will  be  brought  to  rest  and  the 
second  will  take  the  velocity  which  the  first  had. 


[137.]  ELASTICITY.  89 

3d.  Let  the  bodies  be  of  equal  masses,  and  move  in  oppo- 
site directions,  then  M  =  M'  and  v'  will  be  negative,  and 
we  have 


Vi  =  V. 

137.  Impact  of  Imperfectly  Elastic  Bodies.  —  When 
the  bodies  are  imperfectly  elastic  the  force  of  restitution 
will  be  only  the  6th  part  of  the  force  of  compression; 
hence,  the  velocity  due  to  restitution  will  be  the  6th  part 
of  expressions  (1)  and  (2)  of  Article  134,  or 


eM 


and  these,  subtracted  from  the  common  velocity  at  the  in- 
stant of  greatest  compression,  give  the  final  velocity.  The 
results  will  be  the  same  as  equations  (5)  and  (6)  of  Article 
135,  after  the  last  terms  of  those  equations  have  been 
multiplied  by  e  ;  hence,  we  have 

Mv+M'v'          eM'      . 


,     Mv  +  M'v' 


Let  the  body  M'  be  indefinitely  large  and  at  rest,  then, 
making  M!  =  oc  and  v'  =  0,  we  have 


=  —  ev 


90  KINETICS.  [138.] 

hence,  the  former  body  will  rebound,  and  by  disregarding 
the  signs,  using  the  numerical  values  only,  we  have 


which  is  the  same  as  equation  (1)  of  Article  132. 

Multiplying  equation  (1)  by  M  and  (2)  by  M.',  anr1 
adding  the  results,  gives 

Mvj.  4-  M'v\  =  Mv  +  M'v'  ; 

and,  since  the  index  of  elasticity  has  disappeared,  the 
total  momentum  of  the  bodies  before  impact  will  be  the 
same  as  after  impact  ;  or,  in  other  words,  the  total  momen- 
tum of  a  free  system  remains  constant. 

138.  Loss  of  Kinetic  Energy  due  to  Impact.  —  The 
total  kinetic  energy  of  both  bodies,  before  impact,  will  be 


and  is  independent  of  the  directions  of  the  movement  of 
the  bodies.     After  impact  the  kinetic  energy  will  be 


in  which  substitute  the  values  of  v±  and  v\  from  the  pre- 
ceding  article,  and  we  find 


Since  e2  is  always  less  than  unity  for  solid  bodies,  1  —  £ 
will  be  positive,  and  the  last  term  must  be  subtracted  from 
the  preceding;  hence,  practically,  in  the  impact  of  solid 
bodies  there  is  always  loss  of  kinetic  energy.  The  loss  will 
be  equivalent  to  the  heat  developed  by  the  impact. 

Tf  the  restitution  were  perfect,  we  would  have  6=1, 
and  the  expression  would  become 


[138.]  ELASTICITY.  91 

hence,  in  the  impact  of  perfectly  elastic  bodies,  no  energy 
is  lost.  This  result  shows  the  great  utility  of  springs 
under  carriages,  carts,  cars,  etc.,  when  they  are  drawn  over 
rough  roadways.  A  horse  will  do  more  useful  work  by 
drawing  loads  upon  a  cart,  the  body  of  which  is  supported 
by  springs,  than  if  the  cart  were  unprovided  with  springs, 
and  a  locomotive  will  consume  less  coal  in  hauling  a  train 
of  cars  properly  mounted  on  springs  than  it  would  if  there 
were  no  springs  under  them. 

To  find  the  loss  of  kinetic  energy  for  perfectly  non- 
elastic  bodies,  make  e  =  0  in  the  above  equation. 

EXAMPLES. 

1.  If  a  body,  whose  weight  is  20  Ibs.,  is  pulled  by  a  con- 

stant force  of  5  Ibs.  for  5  seconds;  required  the 
momentum  produced. 

2.  A  prismatic  bar  of  iron,  whose  section  is  0.75  of  a 

square  inch,  length  10  feet,  coefficient  of  elasticity 
26,000,000  Ibs.,  is  stretched  by  a  pull  of  9,000  Ibs", 
what  will  be  the  elongation  ? 

3.  A  cylindrical  bar  of  iron,  whose  diameter  is  •£  inch, 

length  2  feet,  is  elongated  -gV  of  an  inch  by  a  pull  of 
2,500  Ibs. ;  required  the  coefficient  of  elasticity. 

4.  Two  perfectly  non-elastic  bodies,  whose  weights  are  10 

and  8  Ibs.,  and  velocities  12  and  15  feet  per  second 
respectively,  moving  in  opposite  directions,  impinge 
upon  each  other;  required  their  common  velocity 
after  impact. 

5.  A  ball,  weighing  20  Ibs.,  moving  with  a  velocity  of  100 

feet  per  second,  overtakes  a  ball  weighing  50  Ibs., 
moving  with  a  velocity  of  40  feet  per  second,  their 
modulus  of  restitution  being  £;  required  their  velo- 
cities after  impact.  Ans.  35-ij-,  and  65^-  feet. 


92  KINETICS.  [138.J 

6.  In  the  preceding  example,  suppose  the  second  body  to 

be  at  rest ;  required  the  velocities  after  impact. 

Ans.  —  7|,  and  +  42f  feet. 

7.  In  Example  5,  suppose  that  they  move  in  opposite 

directions  with  the  velocities  given ;  required  the 
velocities  after  impact. 

Ans.  —  50,  and  +  20  feet. 

8.  A  body  falls  from  a  height  h  upon  &  fixed  plane  of  the 

same  substance  and  rebounds  ;  the  modulus  of  resti- 
tution being  0,  required  the  whole  distance  it  will 
move  in  being  brought  to  rest. 

A          1+g2z 

Ans.  i 5  h. 

1  —  e? 

9.  In  the  preceding  example  find  the  whole  distance  when 

0  =  1,  £,  ^  or  0. 

10.  A  body  impinges  upon  an  equal  body  at  rest ;  show 

that  the  kinetic  energy  before  impact  cannot  exceed 
twice  the  kinetic  energy  of  the  system  after  impact. 

EXERCISES. 

1.  Is  elasticity  a  force  ? 

2.  If  TF  is  the  weight  of  a  body  and  v  its  velocity,  is  We  the  momentum  ? 

3.  Does  momentum  enable  one  to  determine  the  amount  of  resistance 

which  a  moving  body  may  overcome  ? 

4.  Suppose  that  it  is  required  to  determine  how  far  a  ball  would  pene- 

trate a  body  if  fired  into  it ;  would  the  solution  be  effected  by  the 
principles  of  momentum,  work,  energy,  vis  viva,  elasticity,  or  by 
simple  force  ? 

5.  If  two  bodies  move  along  rough  surfaces,  and  finally  impinge  upon 

each  other,  will  the  formulas  of  Articles  133  and  135  enable  one 
to  determine  the  velocities  after  impact  ? 

6.  Can  two  perfectly  non-elastic  bodies  of  unequal  masses  approach 

each  other  with  such  velocities  as  to  destroy  their  motions  ? 

7.  Can  two  perfectly  elastic  bodies  huve  such  relative  masses  and  velo- 

cities that  they  will  mutually  destroy  each  other's  motion  by  th« 
impact  of  one  upon  the  other  ? 


[189.]  FORCE  — ENERGY  — WORK,    ETC.  93 

8.  Is  there  any  relation  between  the  coefficient  of  elasticity  and  the 

modulus  of  restitution  f 

9.  Explain  how  the  use  of  springs  may  prolong  the  life  of  cars,  and 

also  the  track  on  railways. 

Force,  Energy,    Work,  Momentum. 

139.  The  office  which  these  several  elements  perform  in 
the  solution  of  problems  is  best  shown  by  an  example. 
Suppose  that  two  imperfectly  elastic  bodies  impinge  one 
upon  the  other.  That  which  gives  them  motion  is  force. 
That  which  determines  their  velocity  after  impact  is 
momentum.  That  which  determines  their  ability  to  com- 
press, or  break,  or  damage  each  other  is  energy,  and  the 
compression  is  work  done.  The  permanent  compression 
remaining  in  the.  bodies  represents  kinetic  energy  lost, 
which  has  passed  into  an  equivalent  amount  of  heat. 


CHAPTEK  VI. 

COMPOSITION    AND    KESOLUTION   OF    PRESSURES. 

140.  Remark. — A  force  which  acts  upon  a  body  with- 
out producing  motion  results  in  pressure  only.     If  several 
pressures  concur  they  may  evidently  be  replaced  by  a 
single  pressure  which  will  produce  the  same  effect  as  the 
combined  effort  of  all  the  pressures.     The  single  pressure 
is  called  the  resultant  pressure.     It  might,  perhaps,  be 
assumed  that  the  value  of  the  resultant  for  statical  pressures 
is  the  same  as  for  those  pressures  which  produce  motion, 
but  many  think  that  it  is  best  to  deduce  the  value  for  the 
former  independently  of  the  latter,  as  has  been  done  in  the 
following  articles.     "We  will  find,  however,  that  the  result- 
ant for  mere  pressures  is  the  same  as  given  in  Arts.  52  to  58. 

141.  If  a  pressure  acts  directly  opposed  to  the  resultant 
of  all  the  other  forces  in  the  system,  and  of  the  same  inten- 
sity as  the  resultant,  the  system  will  be  in  equilibrium.    The 
resultant  of  an  equilibrated  system  is  zero.     Since,  in  stati- 
cal problems,  there  is  always  equilibrium,  any  one  of  the 
forces  reversed  wiU  be  the  resultant  of  all  the  others. 

142.  Component  Pressures. — In  Fig.  36,  if  R  is  the 

resultant  of  the  pressures  F  and  F^ 
the  latter  are  called  component  pres- 
sures. If  the  resultant  is  zero,  all  the 
forces  may  be  considered  as  compo- 
nents. There  may  be  any  number  of 
fta.  38.  components  or  a  single  resultant  acting 

upon  a  particle. 

143.  If  two  equal  pressures  act  upon  a  particle,  the 


[144-146.]       COMPOSITION    OF    PRESSURES. 


95 


direction  of  the  resultant  will  bisect  the  angle  between 
them. 

For  no  reason  can  be  assigned  why  it  should  be  neare~ 
one  than  the  other  of  the  components. 

144.  //  three  equal  pressures  act  upon  a  particle,  making 
ingles  0/1200  with  each  other,  they  will  be  in  equilibrium. 

For  no  reason  can  be  assigned  why  any  one  should 
prevail  over  the  other  two. 

145.  Parallelogram  of  Pressures. — If  two  pressures, 
acting  on  a  particle,  be  represented  in  magnitude  and 
direction  by  two  straight  lines  drawn  FROM  the  particle, 
and  a  parallelogram  be  constructed  on  these  lines  as 
adjacent  sides,  the  resultant  pressure  witt  be  represented 
in  magnitude  and  direction  by  that  diagonal  of  the  par- 
allelogram which  passes  through  the  particle. 

The  proof  of  this  proposition  is  given  in  two  parts : — 
first,  the  direction  of  the  resultant,  and  second,  its  magni- 
tude. The  following  is  Duchayla's  proof : 

146.  Direction  of  the  Resultant  of  Two  Pressures. 
— First,  let  the  forces  be  commensurable. 


Fio.  37. 


Let  F  and  P  be  two  forces  acting  on  a  particle  at  A  ; 
and  Ac,  Fig.  38,  represent  F  and  Ah  represent  P.  Let 
Act  be  the  common  measure  of  the  forces;  F=  3Aa,  and 
P  =  2Aa  =  %Ad.  On  Aa  and  Ad  construct  a  parallelo- 
gram, which,  in  this  case,  will  be  a  rhombus,  and  the 
direction  of  the  resultant  will  be  the  diagonal  Ae,  for  it 


96  STATICS.  [H7."i 

bisects  the  angle  dAa.  Since  a  force  may  be  considered 
as  acting  at  any  point  in  its  line  of  action,  the  resultant 
may  be  considered  as  acting  at  e,  and  the  force  Aa  be- 
comes transferred  to  de,  parallel  to  Aa,  and  Ad  to  the 
line  ae.  Combining  the  forces  de  and  dh  in  the  same 
manner,  their  resultant  will  be  along  the  line  di,  and  the 
three  forces,  Aa,  Ad,  dh,  may  be  considered  as  acting  at 
* ;  hence,  Ai  is  the  direction  of  the  resultant  of  these 
forces.  Now,  combining  ae  (one-half  of  ai}  with  al>, 
gives  af\  and  efwith  ei  gives  ej,  and  the  four  forces,  Aa, 
ab,  Ad,  dh,  become  transferred  to^' ;  hence,  Aj  is  the 
direction  of  their  resultant.  Proceeding  in  this  way,  we 
find  the  resultant  of  Ac  and  Ah  to  be  in  the  direction  of 
AB,  the  diagonal  of  the  parallelogram  AcJSh.  Similarly, 
the  proposition  will  be  true  for  mP  and  nF,  in  which  m 
and  n  are  integers. 

Secondly,  let  the  forces  be  incommensurable.  A  ratio 
may  always  be  found  which  shall  differ  from  the  true 
ratio  by  less  than  any  assignable  quantity,  and  hence  the 
diagonal  AS  of  the  commensurable  part  will  differ  from 
the  direction  of  the  resultant  by  less  than  any  assignable 
quantity.  But  no  reason  can  be  assigned  why  this  diago- 
nal should  be  on  one  side  of  the  resultant  rather  than  on 
the  other ;  hence,  they  will  coincide. 
147.  THE  MAGNITUDE  of  the  resultant  equals  the  length 

of  the  diagonal  of  the  par- 
allelogram, of  which  the 
adjacent  sides  represent  the 
forces. 

Let  AB  and  AC  repre- 
sent   in     magnitude    and 
Pia-89-  direction     the     respective 

forces,  and  AD  the  direction  of  their  resultant.     Take 


[148,149.]      COMPOSITION    OF    PRESSURES.  97 

A  E  011  AD  produced  backward,  and  of  such  length  as  to 
represent  the  magnitude  of  the  resultant  on  the  same 
scale  as  the  forces  P  and  F.  Then,  the  forces  AB,  A  C, 
and  AE  will  equilibrate.  On  AE  and  AB,  as  adjacent 
sides,  construct  the  parallelogram  ABGE,  and  the  diago- 
nal A  G  will  be  the  direction  of  the  resultant  of  AE  and 
AB. 

Hence,  AC  must  be  in  the  same  straight  line  as  AG, 
and  A  GBD  will  be  a  parallelogram ;  therefore  AD  =  GB. 
But  BG  =  AE\  .'.AE=AD\  hence,  the  resultant  of 
.AC' and  AB  equals  AD  in  magnitude. 

In  Fig.  39  we  have 

A&  =  AB*  +  AC*  +  2A£  .  ACcos, BAG, 
or 

If  =  F*  +  P*  +  ZFPcos  (F,P). 

This  equation  is  of  the  same  form  as  those  in  Articles 
14  and  56. 

148.  Triangle    of  Pressures.  —  Tf  three  concurrent 
forces  are  in  equilibrium  they  may  be  represented  in  mag- 
nitude and  direction  of  action  by  the  three  sides  of  a 
triangle  taken  in  their  order. 

In  Fig.  39,  if  EA  is  a  force  equal  and  opposite  to  the 
resultant  of  the  forces  P  and  F,  and  all  three  act  upon  a 
particle  at  A,  they  will  be  in  equilibrium  ;  and,  according 
to  the  preceding  Article,  if  GA  represent  P  in  magni- 
tude, and  the  direction  be  from  G  towards  A,  and  AB 
represent  F  \n.  magnitude  and  direction,  then  will  BG 
represent  the  equilibrating  force  AE. 

It  should  be  specially  noticed  that  the  sides  of  the 
triangle  do  not  represent  the  position  of  the  forces. 

149.  Conversely,  if  three  forces,  acting  upon  a  parti- 
cle, are  represented  in  magnitude  and  direction  by  the 

5 


98  STATICS.  [  150-152.  J 

three  sides  of  a  triangle,  TAKEN  IN  OKDEK,  they  will  keep 
the  particle  in  equilibrium. 

150.  If  three  forces ',  acting  upon  a  particle,  keep  it  in 

equilibrium,  they  will  be  pro- 
portional to  the  sines  of  the 
angles  between  the  other  two. 

Thus,  from  Fig.  40,  if  the 
forces  P,  F,  R,  acting  on  a 
particle  at  A,  keep  it  in  equili- 
brium, we  have 
P:F:JR  =  AC:  CB  :  AB, 

=  8inA£C:  sin  CAB :  sin  A  CB, 
=  sin  (F,  It)  :  sin  (P,  R)  :  sin  (P,  F). 

151.  Proposition. — If  the  directions  of  three  forces  in 
equilibrium  are  given,  and  the  magnitude  of  one  is  also 
given,  the  magnitudes  of  the  other  two  may  be  found. 

The  truth  of  this  follows  directly  from  the  triangle  of 
forces. 

Generally,  if  three  forces,  P,  F,  It,  acting  on  a  par- 
ticle, keep  it  in  equilibrium,  if  any  three  of  the  quantities 
P,  F,  R,  and  the  angles  which  they  make  with  each  othet 
are  given,  the  remaining  three  quantities  may  be  found, 
PROVIDED  one  of  the  given  quantities  is  a  force. 

For,  the  solution  consists  simply  in  solving  a  plane  tri- 
angle, in  which  the  given  parts  are  a  side  and  any  two  of 
the  remaining  parts  of  the  triangle. 

152.  If  three  forces,  acting  on  a  particle,  keep  it  in  equi- 
librium, they  will  be  proportional  respectively  to  the  sides 
of  a  triangle  formed  by  drawing  lines  perpendicular  to 
the  directions  of  the  action-lines  of  the  forces. 

For,  a  triangle  thus  formed  will  be  similir  to  the  tri- 
angle of  equilibrium. 


[153,154]      COMPOSITION   OP    PRESSURES.  99 

153.  If  three  forces,  acting  in  one  plane  upon  a  RIGID 
BODY,  keep  it  in  equilibrium,  their  action  lines  either  all 
meet  at  a  point,  or  are  all  parallel. 

The  lines  of  action  of  two  of  the  forces  may  meet  in 
a  point,  and  their  resultant  must  pass  through  that  point 
'and  may  replace  the  forces;  but,  since  there  is  equili- 
brium, this  resultant  must  be  equal  and  opposite  to  the 
third  force ;  hence,  the  line  of  action  of  the  third  force 
must  pass  through  the  intersection  of  the  lines  of  action  of 
the  other  two.  If  they  do  not  meet  they  are  parallel. 

154.  The  polygon  of  pressures  and  the  parallelopi- 
pedon  of  pressures  follow  directly  from  the  triangle  of 
pressures,  giving  expressions  similar  to  the  corresponding 
proportions  for  velocity,  as  in  Articles  16  and  17. 


EXAMPLES. 

1.  If  the  angle  between  two  forces  is  right,  what  is  the 

value  of  their  resultant  ?     If  it  is  0°  ?     If  180°  ? 

2.  If  the  forces  are  3  Ibs.,  4  Ibs.  and  5  Ibs.  respectively, 

and  are  in  equilibrium,  required  the  angle  between 
the  forces  3  and  4. 

3.  What  is  the  angle  between  the  forces  when  P=F=R'i 

4.  If  P  -  F=  100  Ibs.,  and  0  =  60°,  find  R. 

5.  If  R  =  P  +  F,  required  6. 

6.  What  will  0  be  when  —  R  =  P  —  F\ 

7.  If  /»=50  Ibs.,  the  angle  (P,  F)=3o°  and  (P,  72) =11 5°, 

what  will  be  the  angle  (F,  R),  and  the  values  of  the 
forces  F  and  R  for  equilibrium  ? 

8.  A  string  7  feet  long  has  its  ends  fastened  at  two  points 

in  a  horizontal  line  5  feet  apart ;  a  weight  of  20  Ibs. 
is  suspended  at  appoint  3  feet  from  one  end  ;  required 
the  tension  on  the  two  parts  of  the  string. 


100  STATICS.  [155,156.] 

9.  Two  forces  represented  by  two  chords  of  a  semicircle 
passing  from  a  point,  show  that  their  resultant  will 
be  represented  by  that  diameter  of  the  circle  which 
passes  through  the  point. 

EXERCISES. 

1.  In  Fig.  39,  the  diagonal  joining  C  and  B  represents  the  resultant  of 

what  forces  ? 

2.  In  the  same  figure  what  will  represent  the  resultant  of  R  and  F  ? 

3.  In  the  same  figure,  if  AD  represents  a  force  acting  away  from  A,  and 

another  equal  force  should  act  along  the  line  BG  in  the  opposite 
direction,  would  they  be  in  equilibrium  ? 

4.  Under  what  conditions  will  two  forces  be  in  equilibrium  ? 

5.  Can  a  particle  be  kept  at  rest  by  three  forces  whose  magnitudes  are 

as  4,  5,  and  9  ?     Or  as  3,  4,  and  8  ? 

6.  If  R  is  the  resultant  of  P  and  F,  will  P  and  F  act  when  R  is  acting  ? 

Resolution  of  Forces. 

155.  The  resolution  of  forces  consists  in  finding  two  or 
more  components  whose  united  action  will  equal  that  of 
the  given  force. 

156.  Rectangular  Components. — Let  R  be   a  force 
whose  components  F  and  P  form  a  right  angle  with  each 
other.    Let 

a  =  the  angle  between  It  and  P, 
/3  =  the  angle  between  R  and  F\ 
then  we  have 

P  =  R  cos  a  ; 

F  =^cosS^y?sina 


Pio.  41. 


the  last  of  which  may  be  found  by  squaring  and  adding 
the  two  former ;  or,  by  observing  that  R  is  represented  by 
the  hypothenuse  of  a  right-angled  triangle,  of  which  the 
sides  represent  P  and  F. 


[157.] 


RESOLUTION    OF    FORCES. 


101 


157.  To  find  the  angles  a  and  /?. — Since  forces  may 
act  at  all  possible  angles,  and  be  applied  at  points  any- 
where in  space,  it  is  desirable  to  have  a  definite  rule  for 
determining  the  angles  which  they  make  with  the  axea 
of  x  and  y. 

In  the  first  place,  draw  a  line  away  from  the  origin  of 


FIG.  42. 


Fio.  43. 


coordinates  O,  parallel  to  the  line  of  action  of  the  force, 
and  in  the  direction  of  action  of  the  force.  If  the  force 
appears  to  act  towards  the  origin  as  AO,  Fig.  43,  it  must 
be  prolonged  so  that  it  may  be  represented  by  OF. 

Then,  secondly ',  conceive  that  the  angle  a  is  generated 
by  a  line  starting  from  the  axis 
OX)  and  revolving  about  O  to  the 
left,  until  it  coincides  with  the 
action-line  of  the  force  ;  the  angle 
thus  generated  will  be  +  a.  In  a 
similar  way,  +  /8  will  be  generated 
by  a  line  revolving  from  the  axis 
OY  about  O,  to  the  right.  Thus, 
in  Fig.  42,  a  is  an  acute  angle,  in 
Fig.  43,  it  is  nearly  360°,  and  in  Fig.  44  it  is  between  90° 
and  180°  ;  and  y3  in  Fig.  42  is  acute,  in  Fig.  43,  between 
90°  and  180°,  and  in  Fig.  44,  between  270°  and  360°.  It 
is,  however,  often  more  convenient  to  measure  the  angle 


Fro.  44 


102  STATICS.  [158,159.] 

negatively  •  thus,  in  Fig.  43,  —  a  is  the  angle  XOf]  and, 
in  Fig.  44,  -£is  TOR 

These  rules  are  arbitrary,  but  a  rigid  observance  of  them 
secures  uniformity  in  practice. 

158.  Problem. — To  find  the  rectan* 
gular  components  of  any  number  of 
concurrent  forces  in  a  plane. 

Let  O  be  the  point  of  concurrent 
action,  and  through  this  point  draw 
two  lines,  OX  and  OY,  at  right  an- 
gles with  each  other.  These  lines,  in 
Analytical  Geometry,  are  called  rec- 
tangular axes. 

Let  JFl,  Fit  F3,  etc.,  represent  the  intensities  of  the  re- 
spective forces  ; 
a*,  03,  as,  etc.,  the   angles  which    the   forces  make 

respectively  with  the  axis  of  x ; 
A)  A?  A?  etc-?  the  corresponding  angle  with  the  axis 

'      of  y; 

JT,  the  sum  of  the  components  along  the  axis  of  a?,  and 
Yj  the  sum  of  the  components  along  y  • 

then,  according  to  Article  156,  we  readily  find 

X  =  FI  cos  «!  +  F2  cos  03  +  Fs  cos  03  +  etc.  =  2  Fvos  a ; 
Y=  Fl  cos  ft  +  F*  cos  &  +  F3  cos  &  +  etc.  =  S^cos  /3 ; 

in  which  the  expression  SFcosa  means  the  sum  of  a 
series  of  terms  of  the  form  ^cos  a. 

It  is  not  necessary  that  the  origin  of  coordinates  be  at 
the  particle  on  which  the  forces  act. 

159.  Resultant  of  any  number  of  concurrent  forces. 

Let  R  be  the  resultant,  then,  according  to  Article  150, 

we  have 

Rl  =  X2  +  Y\ 


1 160,  161. J          RESOLUTION    OF    FORCES.  103 

If  the  given  forces  are  in  equilibrium  among  themselves 

we  have 

lt  =  0  .:  JT=Oand  Y  =  0. 

160.  To  find  the  direction  of  the  resultant  we  readilj 
deduce  from  Fig.  41, 

cos  (X,  X)  =      ;  cos  (B,Y)  =      . 


EXAMPLES. 

1.  Find  the  resultant  of  the  concurrent  forces  in  the  plane 

xy ;  _^  =  20,  «!  =  30°  ;  F2  =  30,  <%=  90°  ;  7^40, 
03  =  150°  ;  and  f\  =  50,  at  =  180°,  and  find  the 
angle  between  R  and  x. 

2.  If  the  forces  F,  =  20,  ^  =  180° ;  Fa  =  lO,al  =  270°, 

are  concurrent,  find  R. 

3.  If  four  forces  are  all  equal  to  each  other  and  concur- 

rent, and  «!  =  0,  02  —  90°,  as  -  225°,  a4  =  270°,  find 
R  and  the  angle  which  it  makes  with  the  axis  of  x. 

161.  If  the  forces  are  referred  to  three 
rectangular  axes,  we  have 

X=  FI  cos  ai+Fz  cos  cta+etc.  =  EFcos  a ; 
Y=  Fi  cos  fii  +  Ft  cos  /32+etc.  =  E^cos  ^3  ; 
Z  =  Fi  cos  71  +  ^  cos  74  +  etc.  =  E^cos  7  ; 


X          .       F  ^ 

=  -;  cos/3  =  -;  0037  =  -. 


Fia.  4«. 


CHAPTER  VII. 


MOMENTS   OF   FORCES. 

162.  Definition. — The  moment  of  a  force  is  a  measure. 
of  its  effect  in  producing  rotation,  or  of  its  tendency  to 
produce  rotation. 

163.  Measure  of  the  Moment. — The  moment  of  a 
force,  in  reference  to  a  point,  is  the  product  of  the  force 
into  the  perpendicular  distance  of  the  action-line  of  the 
force  from  the  point. 

Let  a  particle  w  or  w'y  Fig.  47,  be  connected  with  a 


BTM 


FIG.  47. 


Fio.  48. 


point  0  by  a  line  without  weight,  and  let  a  force  F  act 
at  any  point  A  of  this  line  and  perpendicularly  to  it. 
The  effect  of  the  force  will  vary  directly  as  the  distance  A 
from  the  point  O.  Suppose  that  rotation  has  been  pro- 
duced, as  shown  in  Fig.  48,  the  particle  having  been  moved 
through  an  angle  A  OB.  The  point  of  application  of  the 
force  will  have  moved  over  the  space  AB,  and  the  work 
done  by  the  force  will  be  (Art.  92) 

F.AB. 
If  the  point  of  application  were  at  A',  it  would  have 


[164,165.]  MOMENTS    OF    FORCES.  105 

moved  over  the  space  A'B'  in  moving  the  particle  w  from 
A  to  .B,  and  the  work  done  by  F  would  be 

F.  A'B'. 
But, 

AB'.AB'::OA  :  OA '; 
hence, 

F.AB:  F.  A'B'r.  OA  :  OA' ; 

that  is,  the  effect  of  a  force,  in  producing  rotation,  varies 
directly  as  the  perpendicular  distance  of  the  force  from  the 
point  about  which  rotation  takes  place.  The  effect  evi- 
dently varies  directly  as  the  intensity  of  the  force ;  hence, 
generally,  the  effect  will  vary  as  the  product  of  the  force 
into  the  distance  of  the  action-line  of  the  force  from  the 
point.  This  is  called  the  moment,  as  given  above. 

164.  If  the  line  of  action  of  the  force  is  inclined  to  the 
line  OA,  resolve  the  force  into  two  compo- 
nents, one  F%,  acting  along  the  line  AO, 

the  other,  F^  acting  perpendicular  to  OA 
at  the  point  of  application  A,  of  the  given 
force.  The  former  does  not  tend  to  pro- 
duce rotation  about  O,  but  the  latter  acts  in 
the  same  manner  as  in  Fig.  47.  Hence, 
we  have,  for  the  measure  of  the  moment  in 

FIG.  49. 

this  case, 

Fl.OA=Fsin OAB . OA  =  F. OA  sin OAB  =  F. OB. 
But   OB  is  the  perpendicular  from  0  upon  the  lino 
AB  of  the  force  F. 

165.  Axis  of  Moments. — Rotation   about   a  point  is 
always  equivalent  to  a  rotation  about  an  axis  which  passes 
through  the  point,  and  perpendicular  to   the   plane   of 
motion  of  the  particle.    This  line  is  called  the  axis  of  rota- 
tion, and,  in  reference  to  the  moments  of  forces,  is  the 
axis  of  moments,  or  moment  axis. 

5* 


106  STATICS  [166-1 68.  j 

166.  If  the  axis  of  rotation  is  fixed,  and  the  line  of 
action  of  the  force  is  inclined  to  the  plane  of  motion ;  in 
order  to  find  the  moment  of  the  force,  the  force  is  re- 
solved into  two  components,  one  of  which  is  perpendicu- 
lar to  the  plane  of  motion  and  the  other  parallel  to  it. 
The  former  has  no  moment  in  reference  to  the  fixed  axis,' 
and  the  moment  of  the  latter  will  be  found  by  Article 
164. 

167.  The  point  or  axis  about  which  the  particle  or  body 
rotates  may  not  only  move  in  space,  but  may  also  change 
its  position  within  a  body. 

168.  Definitions. — The  point  O,  where  the  axis  of  rota- 
tion pierces  the  plane  of  motion  of  the  particle,  or  the 
plane  in  which  it  tends  to  move,  is  called  the  origin  of 
moments. 

The  perpendicular  OJ2,  let  fall  from  the  origin  of  mo- 
ments upon  the  action-line  of  the  force,  is  called  the  arm 
of  the  force. 

The  moment  of  a,  force  in  reference  to  a  point  is  the 
product  of  the  force  into  its  arm.  If  a  is  the  arm  of  the 
.force  the  moment  will  be 

Fa. 

The  moment  of  the  velocity  is  the  product  of  the  velo- 
city into  the  perpendicular  from  the  origin  of  moments 
upon  the  direction  of  motion.  The  direction  of  motion 
will  be  in  a  tangent  to  the  path  at  the  point  where  the 
velocity  is  considered.  Let  p  be  the  perpendicular,  then 
the  moment  of  the  velocity  will  be 

pv. 

The  moment  of  the  momentum  is  the  continued  product 
of  the  mass,  velocity,  and  perpendicular  from  the  origin  of 


[169-171.]  MOMENTS    OF    FORCES.  107 

moments  upon  the  direction  of  motion.     Let  Q  be  the 
momentum,  then 

Qp  =  Mvp. 

This  principle  is  important  hi  the  solution  of  certain 
problems  involving  an  aggregation  of  particles. 

The  moment  of  a  force  oblique  to  the  axis  of  rotation 
is  the  product  of  the  component  of  the  force  on  a  plane 
perpendicular  to  the  axis  into  the  arm  of  the  component. 

Generally,  the  moment  of  a  mechanical  agent  is  a  meas- 
ure of  its  importance  in, producing,  or  tending  to  pro- 
duce, rotation. 

169.  The  moment   of  a  force  may  be  represented   by 
twice  the  area  of  a  triangle,  of  which  the  base 
represents  the  magnitude  and  position  of  the 

force,  and  whose  apex  is  at  the  origin  of  mo- 
ments.    For,  the  altitude  of  the  triangle  will 

be  the  perpendicular  upon  the  force,  and      L jp~ 

hence,  will  be  the  arm  of  the  force. 

170.  Sign  of  the  Moment. — If  a  force  tends  to  turn 
a  system  one  way,  it  may  be  considered  positive  ;  then,  if 
in   the  opposite   direction,   it   will  be   negative.     Either 
direction  may  be  considered  positive,  but  when  chosen  it 
must  not  be  changed  in  the  solution  of  a  problem. 

If  a  watch  be  placed  at  the  origin  of  moments,  with  its 
face  in  the  plane  of  the  force,  the  moments  of  those  forces 
which  tend  to  turn  the  particle  or  body  in  the  direction  of 
the  movement  of  the  hands  of  the  watch  will  be  right- 
handed,  and  in  the  opposite  direction  left-handed. 

171.  The  value  of  a  moment  may  be  represented  in 
magnitude  and  direction  by  the  axis.     When  the  moment 
is  positive,  let  the  axis  be  represented  above  the  plane,  aa 
in  Fig.  51,  and  a  distance  laid  off  on  it  to  represent  its 


108 


STATICS. 


[172.] 


magnitude.     "When  the  moment  is  negative,  lay  off  the 
axis  below  the  plane  of  the  force,  as  in  Fig.  52. 


-Fa 


Fio.  51. 

172.  Composition  of  Moments. — Since  moments  are 
fully  represented  by  lines,  they  may  be  compounded  or 
resolved  in  the  same  manner  as  forces.  Thus,  if  the 
forces  are  all  in  the  same  plane,  their  axes  will  be  paral- 
lel, and,  if  they  have  the  same  origin,  their  axes  will 
coincide ;  in  which  case,  the  resultant  moment  will  be  the 
algebraic  sum  of  the  several  moments. 

If  Or  be  the  resultant  moment,  we  have 

Or  =  Fa  +  ^I«i  +  ^2«2  +  etc. 

If  the  forces  act  in  different  planes  let  all  their  moment 
axes  pass  through  A,  Fig.  53, 
and  let 

GI  =  JF[ai  =  the    moment    of 

one  force, 
O2  =  F-fa  =  the    moment    of 

another  force, 

0  =  the  angle  between  the  mo- 
ment axes,  and 
Or  =  the  resultant  moment  of 

Oi  and  0% ; 
then,  as  in  Article  147,  we  have 

O*  =  O?  +  6>22  +  26^2  cos  0. 
If  the  axes  do  not  pass  through  a  common  point,  it  maj 


[  173, 174] 


MOMENTS    OF    FORCES. 


109 


still  be  proved  that  the  resultant  moment  may  be  found 
from  the  preceding  equation. 

173.  Proposition.  —  If  any  number  of  concurring  forces 
are  in  equilibrium,  the  algebraic  sum  of  their  momenta 
will  be  zero. 

Let  jF[,  ^2,  F3,  etc.,  be  the 
forces  acting  upon  a  particle  at 
A.  Take  the  origin  of  moments 
at  any  point  O.  Draw  OA,  and 
let  fall  the  perpendiculars  Oa, 
Ob,  Oc,  etc.,  upon  the  action-lines 
of  the  forces  ;  and  let 
Oa  =  a>i;  6$  =  #2  ;  Oc  =  as,  etc. 

Since  they  are  in  equilibrium, 
the  sum  of  the  components  of  all  the  forces  perpendicular 
to  OA  (see  Article  142)  will  be  zero  ;  hence, 

t  +  F2sinOAF3  +  F98mOAF3  +  etc.  =  0; 


FIQ.  64. 


or 


,.,  Oa 
^  OA 


+FS 


Oc 
OA 


+  etc.  =  0. 


Multiplying  by  OA,  we  have 


.0c  +  etc.  =  0; 
etc.  =  2Fa  =  0. 


or 


When  there  is  equilibrium  in  reference  to  rotation,  any 
one  of  the  moments  may  be  considered  as  equal  in  value 
but  directly  opposed  to  the  resultant  of  the  moments  of 
all  the  other  forces. 

]/74.  Unit  of  Moments.  —  If  the  force  be  given  in 
pounds,  and  the  arm  in  feet,  the  unit  will  be  a  foot-pound, 
and  the  moment  will  be  a  certain  number  of  foot-poundf 


110 


STATICS. 


[  175-177.  J 


of  rotary  effort.    It  will  be  observed  that  this  is  not  the 
game  as  foot-pounds  of  work. 

175.  The  origin  of  moments  may  ~be  taken  anywhere 
in  the  plane  of  the  forces.     This  is  evident  from  the  pre- 
ceding article ;  for,  the  point  O,  in  Fig.  54,  was  not  only 
chosen  arbitrarily,  but  no  trace  of  its  position  remains  in 
the  result.     This  is  also  evident  from  the  fact  that  if  the 
forces  are  in  equilibrium  their  tendency  to  turn  the  body 
about  any  point  is  zero.     In  statical  problems,  therefore, 
the  origin  may  be  chosen  arbitrarily. 

176.  The  arm  of  a  force  in  terms  of  Rectangular 
Coordinates. — Let  F  be  a  force,  of  which  Aa  is  its  line 

of  action.  Take  the  origin  of 
coordinates  at  0,  which  is  also 
taken  as  the  origin  of  mo- 
ments ;  then  Oa  will  be  the 
arm  of  the  force. 

—  X         Take  any  point  A,  in   the 
line  of  the  force,  and  drop  the 
perpendicular  Ab,  and    from 
its  foot  drop  the  perpendicular 
bo  upon  aA,  and  draw  Od  parallel  to  a  A. 

Let  a  =  dOb  ;  (3  =  the  angle  between  ^and  Y  —  cAb. 
y  =  Ab  ;  x  =  Ob. 

Then 

cb  =  y  cos  a  •  db—x  cos  ft ; 

.".  aO  —  cb  —  db  =  y  cos  a  -  x  cos  /3. 

177.  The  origin  of  coordinates  may  be  at  one  place  and 
the  origin  of  moments  at  any  other,  for  the  positions  of 
both  are  arbitrary.     Thus,  in  Fig.  55,  the  origin  of  mo- 
ments may  be  taken  at  d,  or  &,  or  any  other  place,  the 
origin  of  coordinates  remaining  at  O. 


0 


Fio.  65. 


[178,179.]  MOMENTS    OF    FORCES.  Ill 

178.  Moments  of  Parallel  Forces. — If  the  forces  are 
parallel  their  arms  will  coincide.  Thus,  in  Fig.  56,  if  O 
be  the  origin  of  moments,  then  will  the  arms  of  the  forces 
FI,  F2,  Fs,  etc.,  be  Oa,  Ob,  Oc,  etc.,  respectively. 


d       > 

C                     y 

I 

* 

a 

O 

Fia.  56. 

179.  Problems.  —  1.  A  weight  W  is  suspended  from  one 
end  of  a  horizontal  bar  A  C,  the  other  end  of  which  rests 
against  a  vertical  wall,  the  bar  being  held  in  position  by 
a  cord  DB  j  required  the  tension  of  the  cord. 

Take  the  origin  of  moments  at  A,  where  the  bar  touches 
the  wall.  The  perpendicular  AE  upon  the  cord  BD  will 
be  the  arm  of  the  tensile  force  of  the  cord,  and  A  C  will 
be  the  arm  of  W.  Let  t  be  the  tension  of  the  cord,  then 
will  t.AJE  be  the  moment  of  the  tension,  and  we  have 

t.AE=W.AC. 


This  problem  illustrates  the  mechanical  arrangement  of 
the  bones  and  muscles  at  the  elbow-joint,  by  means  of 
which  the  arm  may  be  held  in  a  horizontal  position  and 
support  a  weight  in  the  hand.  The  joint  is  at  A,  the 
hand  at  C\  and  the  muscle  at  DB  ;  but  in  the  arm  the 
distance  AE  is  much  less  in  proportion  to  A^}  than  that 
shown  in  the  figure. 

2.  A  weight  W  is  suspended  by  a  cord  2)£,  and  pushed 


112  STATICS.  [180,181.] 

from  a  vertical  by  a  bar  AB  •  required  the  tension  of 

the  cord. 

Take  the  origin  of  moments  at  A,  and  drop  the  perpen- 

diculars AC  and  BE.  The  perpendicular  BE  will  be 
of  the  same  length  as  if  it  were  drawn 
horizontally  from  A  to  a  vertical 
through  B.  If  t  be  the  tension,  we 

have 

t.AC=W.BK 


AC 

If  the  length  and  inclination  of  AB  be  given,  we  have 
BE=.  AS  sin  0. 

If  the  length  of  DB  be  also  known,  the  angle  ADB  =  $ 
may  be  found.  Or,  if  the  three  lines  AD,  AJB,  DJB,  are 
given,  the  angles  <j>,  0,  ABD,  may  be  found  by  solving  the 
triangle. 

180.  Choice  of  the  Origin  of  Moments.  —  When  the 
forces  are  in  one  plane,  the  problem  may  be  solved  by  a  sin- 
gle equation  of  moments,  provided  the  origin  of  moments 
can  be  so  taken  that  the  moment  of  one  force  only  will  be 
unknown.     It  will  be  observed  that  the  forces   at   the 
origin  of  moments  have  no  moments,  and  hence  do  not 
enter  the  equation  of  moments.     When  there  are  several 
unknown  forces,  it  is  possible,  in  many  cases,  to  find  the 
value  of  all  of  them  by  taking  the  origin  of  moments  at 
different  places,  so  as  to  involve  only  one  unknown  quan- 
tity at  a  tiifte. 

181.  Problems.  —  1.  In  Fig.  58,  suppose  that  the  bar  AB 
rests  against  a  smooth  surf  ace  DA,  and  is  prevented  from 


M81.]  MOMENTS    OF    FORCES.  113 

sliding  upward  by  a  string  AE  ;  required  the  tension  of 
the  string  and  the  pressure  against  the  surface. 
Let 

F=  the  tension  on  A£\  and 
X  =  the  pressure  against  the  surface. 

If  the  origin  of  moments  be  taken  at  B,  both  the  un- 
known forces  jPand  X.  will  enter  the  equation,  and  hence, 
neither  will  be  determined.  If  it  be  taken  at  D  the  mo- 
ment of  F  will  be  zero.  The  forces  F  and  X  are  the 
components  of  the  compression  along  BA  ;  hence,  the 
latter  will  not  be  considered  when  the  former  are. 
Taking  the  origin  of  moments  at  D,  we  have 


.  x-  —  W 
~  DA  F 

Now,  taking  the  origin  of  moments  at  J?,  there  results 
F.BE=X.AE; 

in  which,  substitute  the  value  of  X  from  above  and  reduce, 
and  find 


The  compression  along  AB  will  be 


DA  ~  DA  " 

The  compression  may  be  found  directly  by  taking  the 
origin  of  moments  at  D.  Drop  a  perpendicular  from  D 
upon  BA  prolonged  (which  the  student  can  draw  in  the 


114 


S'l  A.TICS. 


[181.] 


figure),  and  call  it  a.    Let  e  =  the  compression  ;  then  the 
equation  of  moments  will  be 


but 


a  =  DA  sin  e  =  DA 


EB 
AB 


a 


DA 


as  before. 

2.  Two  forces,  P  and  F,  act  as  a  putt  upon  the  ends 
of  a  bar  AB,  both  of  which  are  inclined  to  the  bar  at 
known  angles,  the  point  G  being 
fixed,  and  the  distance  CB 
known;  required  the  distance 
AC  for  equilibrium. 

Take  the  origin  of  moments  at 
C,  and  drop  the  perpendiculars 
Ca  and  Ob  ;  then  the  equality  of 
moments  gives  the  equation 


FIG.  69. 


The  distance  Cb  is  known  from  the  equation 


and  A  C  from  the  equation 

A  C  =  Ca  cosec  a  A  C. 

Substitute  in  this  equation  the  value  of  Ca  found  from 
the  first  equation,  and  Cb  from  the  second,  and  we  find 


.>  —  --.  -T—  TL 

P  sm  a  A  C 
The  resultant  of  the  forces  P  and  F  will  pass  through 


[181.J  MOMENTS    OF    FORCES.  115 

their  intersection  D,  and  also  through  the  fixed  point  C. 
The  moment  of  the  resultant,  when  the  origin  is  at  any 
point  upon  i^  will  be  zero.  Its  value,  however,  may  be 
found  by  taking  the  origin  of  moments  at  A  or  B. 

3.  A  weight  Pv  is  made  to  act  vertically  upward  at  the 
end  of  a  rigid  bar  OA,  and  P%  vertically  downward  at  B, 

on  the  same  bar,  the  bar  being 
free  to  turn  about  the  end  O  ; 
required  the  distance  Ob  for 
equilibrium. 

Take  the  origin  of  moments 
FlG7(5o.  at  O,  and  draw  the  horizontal 

line    Oc,   cutting  the  verticals 

through  B  and  A  at  b  and  c  respectively ;  then  will  the 
equation  of  moments  give 


If  the  distance  be  between  the  weights  is  given,  then 

Oc  =  Ob  +  be  ; 
which,  in  the  preceding  equation,  gives 


f* 

EXAMPLES. 

1.  In  Fig.  57,  if  W  =  20  Ibs.,  AC=<2  ft.,  AS  =  6  in., 

and  AD  —  4  in.,  find  the  tension  on  the  cord  DB. 

2.  In  the  preceding  example  find  the  horizontal  pressure 

against  the  wall. 

3.  In  Fig.  57,  if  the  surface  at  A  is  perfectly  smooth,  find 

the  vertical  force  F  applied  at  A,  which  will  just 
prevent  sliding. 


STATICS.  [182.] 

4.  If  the  cord  J)J3,  Fig.  57,  is  inclined  45°,  what  must  be 

the  distance  from  A  to  B  that  the  tension  on  the 
string  shall  equal  the  weight  W  \ 

5.  In  Fig.  58,  if  DB  =  2A£,  0  =  45°,  and  W=  50  Ibs., 

required  the  tension  on  the  cord  and  the  compres- 
sion on  the  bar. 

6.  A  strut  JBC\  free  to  turn  about  its  lower  end,  supports 

a  weight  W  from  its  upper  end, 
the  strut  being  held  by  a  cord  A  G, 
one  end  of  which  is  attached  to  the 
upper  end  of  the  strut  and  the 
other  to  a  point  A  in  the  horizontal 
plane  ;  required  the  tension  of  the 
cord. 

7.  In  the  preceding  example,  what  condition  must  be  ful- 

filled that  the  tension  of  the  cord  will  equal  the 
weight  W\ 

8.  In  Fig.  61,  if  W=  500  Ibs.,  AB  =  6  ft,  BD  =  ±  ft, 

and  D  C  =  8  ft.,  what  will  be  the  compression  upon 


9.  In  Fig.  60,  if  A  =  l\  and  be  —  2  ft.,  what  will  be  the 

distance  Ob  for  equilibrium  ? 

10.  In  Fig.  60,  if  O  I  -  2  ft.,  and  Pl  =  Pz,  what  will  be 
the  distance  bo  for  equilibrium? 

Couples. 

182.  Definition.  —  Two  equal  parallel  forces,  acting  in 
contrary  directions  and  whose  lines  of  action  are  not  coin- 
cident, constitute  a  couple. 

The  last  equation  of  the  preceding  article  is 

Ob  =  --  be; 


1 183,  184.]  COUPLES.  117 

in  which,  if  Pz  =  Pv,  it  becomes 
6tf  =  oo; 

that  is,  in  order  that  two  such  forces  shall  be  in  equili- 
brium, in  reference  to  rotation,  the  origin  of  moments 
must  be  at  an  infinite  distance  from  the  forces ;  in  other 
words,  two  equal  parallel  forces,  whose  directions  are  con- 
trary and  lines  of  action  not  coincident,  cannot  be  in 
equilibrium  in  reference  to  rotation.  Such  a  system  has 
received  a  special  name,  called  a  couple. 

183.  The  office  of  a  couple  is  to  produce,  or  to  tend 
to  produce,  rotation  only.    For,  the  only  other  effect  which 
it  can  produce  is  that  of  translation  ;  but,  since  the  forces 
are  equal  and  directions  contrary,  whatever  translation  is 
produced  by  one  force  in  any  direction  will  be  exactly 
neutralized  by  the  other  in  the  opposite  direction  ;  hence, 
they  cannot  produce  translation. 

184.  Moment  of  a  Couple. — In  Fig.  62,  let  O  be  the 
origin  of  moments,  and  P  at  A  equal  P  at  B,  one  mo- 
ment being  right-handed  and  the 

other  left-handed.      The   moment  JJ* 

will  be 

P.  OB— P.  OA  (not  equal  0).        < 
But, 

OB-OA^AB-,  'FIG.  68. 

hence,  the  preceding  expression  becomes 

P.OA  +  P .AB-  P.OA-9 
or,  simply 

P.AJB-, 

that  is,  the  moment  of  a  couple  is  the  product  of  one  of 
the  forces  into  the  perpendicular  distance  between  the 
lines  if  action  of  the  forces. 

This  is  independent  of  the  origin  of  moments.     If  the 


JL 

I 


118  STATICS.  [185-187.] 

origin  be  at  A,  upon  the  line  of  action  of  one  of  the  forces, 
the  moment  of  the  couple  will  be  the  same  as  the  moment 
of  one  of  the  forces. 

185.  A  couple  can  be  equilibrated  only  by  an  equiva- 
lent couple  having  a  contrary  moment.      For,  the  only 
effect  being  rotation,  such  a  system  of  forces  must  be  em- 
ployed as  will  produce  a  contrary  rotation,  and  this  requires 
an  equivalent  couple. 

186.  A  resultant  couple  is  one  which  will  produce  the 
same  effect  as  the  several  couples. 

If  PI,  P2,  etc.,  be  the  forces  of  several  couples  all  in 
one  plane, 

«x,  #2,  ete-j  be  their  respective  arms, 
7?,  one  force  of  a  resultant  couple,  and 
r,  the  arm  of  the  resultant  couple ; 
then 

RT  ==  P!#I  4-  Ptfh.  +  P?fh  +  etc. 

If  the  couples  are  in  equilibrium,  any  one  of  them  may 
be  taken  as  equal  but  contrary  to  the  resultant  of  all  the 
others. 

187.  Proposition. — If  two  couples,  having  equal  mo- 
ments, but  whose  directions  of  action  are  contrary,  act  upon 

a  body,  they  witt  equilibrate  each  other. 
This  is  evident  from  Article  185,  but  the 
proposition  is  presented  here  in  order  to 
show  that  the  forces  constituting  the 
couples  may  be  applied  at  any  point  of 
the  body,  and  that  the  arms  of  the  con- 
pies  need  not  be  parallel. 

Conceive  that  any  point  O  in  the  body 
is  fixed,  and  taken  as  the  origin  of  moments,  then  will  the 
moments  of  the  couples  in  reference  to  this  point  be 
P!  .  ab  —  Pz .  cd ; 


[188,189.]  COUPLES.  119 

but,  since  their  moments  are  assumed  to  equal  each  other 
we  have 

Pl .  ab  -  Pz .  cd  =  0  ; 

hence,  there  is  no  tendency  to  turn  about  O.  The  same 
may  be  shown  in  reference  to  any  other  point  of  the  body ; 
hence,  the  body  will  be  in  equilibrium  in  reference  to 
rotation. 

188.  Proposition. — A  force,  acting  at  any  point  of  a 
body,  is  equivalent  to  an  equal  parallel  force  at  the  origin 
of  moments,  and  a  couple  whose  mo- 
ment is  the  moment  of  ike  original 

force. 

Let  the  force  P  be  applied  at  A, 
and  the  origin  of  moments  be  at  B.  >  , 

At  B  introduce  two  equal  and  oppo- 

,.  ,  n  i  ,  Fl<»-  64. 

site  forces,  each  equal  and  parallel  to 

the  original  force  P.  Since  the  two  forces  at  B  neutral- 
ize each  other,  the  effect  of  the  three  forces  will  be  the 
same  as  that  of  the  single  force  P.  But,  by  making  a 
new  combination  of  the  forces,  we  have  the  force  acting 
down  at  A,  combined  with  the  equal  parallel  force  at  B 
acting  upward,  constituting  a  couple  whose  arm  is  AB, 
and  the  force  P  at  B  acting  downward. 

When  a  body  is  free  to  move,  a  single  force  acting  upon 
it  may  produce  both  rotation  and  translation,  and  it  may 
be  shown  that  it  will  produce  both,  unless  the  line  of  action 
of  the  force  passes  through  the  centre  of  the  mass  of  the 
body. 

189.  Proposition. — When  several  forces  have  a  result- 
ant, the  sum  of  their  moments  will  be  zero  when  the 
origin  of  moments  is  upon  the  line  of  the  resultant,  for 
the  moment  of  the  resultant  will  be  zero. 


120  STATICS  [100.] 

Three  Parallel  Forces. 

190.  The  relation  between  three  parallel  forces  in  equi- 
librium may  be  found  by  means  of  the  principles  of  mo- 
ments, and  the  result  may  be  extended  to  any  number  of 
parallel  forces. 

In  Fig.  65,  let  the  forces  R,  F,  P,  act  in  parallel  lines 
and  be  in  equilibrium. 


Tr 


FIG.  65.  FIG.  66. 

Taking  the  origin  of  moments  at  D,  we  have 

F.Z>Cy=P.DE', 
and  if  the  origin  of  moments  be  at  E,  we  have 

F.CE=  R.DE. 
Adding  these  equations  gives 

F(DC+CE}  =(. 
but, 

which,  in  the  preceding  equation,  gives 

hence,  the  force  JF]  acting  in  one  direction,  equals  the  sum 
of  the  two  forces  acting  in  the  contrary  direction. 

If  ^and  P  are  given,  we  have,  by  transposition, 
F-  P  =  E. 

The  same  principles  apply  to  Fig.  66,  in  which  the 
directions  of  P  and  F  are  the  reverse  of  those  in  Fio-.  65. 


[191,192.]  PABALLEL    FORCES.  1SI 

By  transposing  R  in  the  last  equation  we  have 

F-P-JS=0; 

in  which,  if  F  be  called  the  typical  force,  and  the  alge- 
braic signs  be  understood,  we  may  write  it 


and  this  expression  is  true  for  any  number  of  parallel 
forces. 

19L  A  rigid  body,  being  acted  upon  by  any  number  of 
parallel  forces  in  one  plane,  it  is  necessary  and  sufficient 
for  equilibrium  that  we  have 

Sib  =  0  ; 


the  former  of  which  will  determine  equilibrium  in  refer- 
ence to  rotation,  and  the  latter  in  reference  to  translation. 

192.  A.  singfe  force  and  a  single  couple  in  one  plane  are 
equivalent  to  a  single  force  equal 

and  parallel  to  the  original  sin-  X"-2* 

gle  force,  but  having  another 
point  of  application. 

If  they  are  parallel,  as  in  Fig. 
6T,  the  resultant  of  the  forces  in 
reference  to  translation  will  be  Fio.^77 

F+  P  -  P  =  F. 

Call  this  resultant  F',  to  distinguish  it  from  the  F  in 
the  figure.  A  force  equal  and  opposite  to  F',  acting  at 
some  point  1),  will  produce  the  couple  F'—DC—F,  which, 
for  equilibrium,  must  be  equal  and  opposite  to  P—AB—P. 
Hence,  to  find  DC,  we  have 

F.DC=P.AB 
p 

•    '  •*-'   V  7TT 

6  F 


122  STATICS.  [193,  196,  j 

'  If  the  forces  are  not  parallel,  combine  F  with  P,  and 
their  resultant  with  —  JP,  and  the  same  result  will  be 
obtained. 

193.  Remark.  —  The   principle'  of   moments,   strictly 
speaking,  is  applicable  only  to  problems  involving  extended 
masses ;  for,  although  we  speak  of  the  moment  of  a  force 
in  producing  the  rotation  of  a  particle,  yet,  in  order  to 
realize  it,  it  is  necessary  to  assume  that  the  particle  is 
connected  with  the  point  about  which  rotation  takes  place, 
by  means  of  a  rigid  bar,  which  is  itself  a  finite  body  and 
not  a  particle. 

194.  Proposition. — A  system  of  forces  acting  in  one 
plane,  if  not  in  equilibrium,  must  be  equivalent  to  a  sin- 
gle force  or  to  a  couple. 

For,  the  resultant  of  two  forces  may  be  found,  and  the 
resultant  of  that  resultant  and  a  third  force,  and  so  on, 
and  if  their  lines  of  action  tKus  intersect  each  other  a  sin- 
gle resultant  may  be  found  ;  otherwise  a  single  couple 
may  be  found. 

195.  Proposition. — A  system  of  forces  in.  one  plane, 
acting  on  a  rigid  body  will  be  in  equilibrium  if  the  alge- 
braic sum  of  the  moments  of  the  forces  vanishes  in  refer- 
ence to  three  points  in  the  plane  not  in  a  straight  line. 

For,  if  they  are  not  in  equilibrium  in  reference  to  rota- 
tion, the  algebraic  sum  of  the  moments  could  not  vanish 
for  any  point  in  the  plane;  and  if  they  had  a  single 
resultant,  then  the  moments  would  vanish  only  for  points 
on  the  line  of  the  resultant. 

196.  Problems. — 1.  A  prism  AF  is  acted  upon  by  a 
couple  in  the  plane  of  the  upper  base  /  required  the  value 
of  the  couple  in  the  plane  of  the.  lower  base  that  will 
equilibrate  the  former. 

The  forces  in  the  plane  of  the  upper  base  tend  to  turn 


PARALLEL    FORCES. 


123 


the  prism  about  an  axis  perpendicular  to  the  base  ;  simi- 
larly the  couple  in  the  lower  base  will  tend  to  turn  it  about 
the  same  axis  ;  hence,  if  their  moments  are  equal  and 
contrary,  they  will  equilibrate  each  other. 

This  shows  that  couples  of  equal  moments  in  paraLel 
planes  are  equivalent. 

The  couples,  in  this  case,  twist  the  body  upon  which  they 


FIG.  69. 

act.  This  effect  is  called  torsion.  The  amount  of  torsion 
depends  upon  the  properties  of  the  material,  as  well  as 
upon  the  size  of  the  body  and  the  moment  of  the  couple. 
The  properties  of  materials  are  investigated  in  works  upon 
the  Resistance  of  Materials. 

A  single  force  applied  at  the  end  of  a  lever,  as  in  Fig. 
69,  will  not  only  twist  the  bod}7,  but  will  push  it  side- 
wise.  For,  as  we  have  seen  in  Article  188,  it  will  be  equi- 
valent to  a  couple  whose  moment  is 

P.AB, 

and  a  force 

P, 

applied  at  A,  the  former  of  which  twists  the  body  and  the 
latter  pushes  it  sidewise.  This  may  be  easily  illustrated 
by  the  student  in  a  variety  of  ways,  such  as  turning  an 
auger  by  one  handle,  twisting  a  long  rod  by  means  of  a 
single-handed  wrench,  etc. 


124 


STATICS. 


[196.', 


2.  A  door,  gate,  or  frame,  supported  by  two  hinges, 
carries  a  weight  ;  required  the  pressure  upon  the  hinges. 

When  hinges  are  employed, 
they  may  be  so  set  that  one  or 
the  other  will  carry  all  the 
vertical  pressures.  In  Fig.  70, 
there  being  no  provision  for  car- 
rying any  of  the  vertical  pres- 
sures at  the  upper  bearing,  they 
will  all  be  supported  at  the  lower 
end. 

Taking  the  origin  of  moments  at  E,  we  have 


•  H-         W 
~  BA 

Similarly,  taking  the  origin  at  A,  we  have 


-DCW- 
'-  ' 


hence, 

H=H,; 

and  as  they  are  parallel  they  constitute  a  couple.  The 
only  remaining  force  is  the  vertical  one  at  B,  and  is  called 
V.  This  force,  combined  with  TT,  must  constitute  the 
equilibrating  couple  ;  hence, 

V=W. 

The  total  pressure  at  the  lower  end  is  the  resultant  of 
Trand  HI,  and  hence  is 


BA* 


[196.]  PABALLBL    FORCES.  125 

EXAMPLES. 

1.  If  three  forces  are  represented  in  magnitude, position, 

and  direction  of  action  by  the  sides  of  a  triangle 
taken  in  their  order,  show  that  they  are  equivalent 
to  a  couple. 

2.  In  Fig.  71,  if  the  forces  act  along  the  sides  of  a  trian- 

gle, P  from  B  toward  C,  F  from  G  toward  A,  and 
R  from  B  toward  A ;  show  that  for  equilibrium  in 
reference  to  rotation,  R  =  P  +  F. 

3.  Equal  weights  are  suspended  at  the  corners  of  a  trian- 

gle ;  required  the  point  where  the  triangle  must  be 
supported  that  there  will  be  equilibrium. 


t 

V 

Fio.  71.  Fio.  72. 

4.  Two  men  carry  175  Ibs.  between  them  on  a  pole,  resting 

on  one  shoulder  of  each ;  the  weight  is  twice  as  far 
from  one  as  from  the  other  ;  how  much  weight  does 
each  carry,  neglecting  the  weight  of  the  pole  ? 

5.  If  a  prismatic  block  of  stone,  whose  width  AB  is  2  ft., 

height  AC  is  3  ft.,  weighs  500  Ibs.,  and  it  be  con- 
sidered that  the  whole  weight  acts  at  the  centre  g, 
what  force,  acting  horizontally  at  G,  will  just  turn 
the  block  about  the  edge  B  \ 

6.  A  man,  whose  weight  is  175  Ibs.,  desires  to  raise  a  body 

which  weighs  4,000  Ibs.  by  means  of  a  lever  8  feet 
long ;  one  end  of  the  lever  being  placed  under  the 


126  STATICS.  [196.] 

body,  how  far  from  the  end  shall  the  fulcrum  be 
placed  BO  that  his  weight  at  the  other  end  shall  just 
balance  the  body  ? 


EXERCISES. 

1.  The  force  being  given  in  pounds,  and  the  arm  in  feet,  is  it  proper  to 

say — a  moment  of  a  certain  number  of  foot-pounds  ?  Foot- 
pounds of  what  ? 

2.  What  is  the  meaning  of  foot-pounds  of  work  ? 

3.  What  is  meant  by  foot-pounds  of  momentum  per  second  ? 

4.  If  velocity  is  given  in  feet,  and  the  arm  also  in  feet,  whan  will  be  the 

unit  of  the  moment  of  momentum  ? 

5.  Can  a  single  force  acting  upon  a  rigid  body  produce  rotation  if  there 

is  not  a  fixed  point  in  the  body  ? 

6.  Will  two  couples  acting  upon  a  rigid  body  in  planes  at  right  angles 

with  each  other  produce  translation  ? 

7.  If  a  person  supports  a  weight  of  100  Ibs.  suspended  from  a  rod  upon 

his  shoulder,  and  he  pulls  down  upon  the  rod  with  a  force  of  25  Ibs. 
with  his  hands  so  as  to  balance  the  weight,  how  much  more  than 
his  own  weight  will  be  the  pressure  of  his  feet  upon  the  earth.  ? 

8.  If  a  hole  be  bored  by  an  auger,  is  the  resistance  to  the  cutting 

equivalent  to  a  couple  ? 

9.  In  Fig.  70,  will  the  pressure  H  at  A  be  increased  if  the  weight  W 

be  placed  at  a  greater  distance  from  I)  ? 


CHAPTER  VIII. 

PARALLEL   FORCES. 

197.  Parallel  Forces  are  such  as  act  along  parallel  lines, 
They  may  be  conceived  as  concurring  in  a  point  at  an  in- 
finite distance,  but  this  would  be  equivalent  to  saying  that 
they  do  not  actually  concur.     They  are  forces  not  acting 
upon  a  single  particle,  but  upon  the  several  particles  of  a 
body. 

198.  The  Resultant  of  Parallel  Forces.  —  Let  B  be 
the  resultant  of  the  parallel  forces  F1}  F2,  etc.  ;   then, 
according  to  Article  191,  we  have 

E  =  fi+ft  +  Fi  +  ete.  =  SF.  '  *' 


It  will  be  observed  that  the  value       —  *  -  ^&3 
of  the  resultant  is  independent  of  the       —  •  -  ^ 
points  of  application   of   the  forces.      "^ 
If  the  forces  are  in  the  plane  xy,  and 
are  resolved  parallel  to  the  axes  x  and  y,  we  have,  accord- 
ing to  Article  156, 

X=  (Fl+Fz+F'3+  etc.)  cos  a  =  cos  a%F\ 
Y=(Fi+F%  +  Ft  +  etc.)  cos  ft  =  cos  02  F=  sin  a 
Squaring  and  adding,  gives 

s2a  +  sin2  a) 


which  is  the  same  as  given  above. 


128  STATICS.  [199,200.] 

If  the  given  forces  are  in  equilibrium,  we  have 

72  =  0; 
.:SF=0;  X=0;   Y=0. 

199.  The  expression  for  the  moments  of  parallel  forces 
is  given  in  Article  178,  and  the  condition  for  equilibrium 
in  reference  to  rotation,  in  Article  191.    These  expressions 
are  independent  of  the  points  of  application  of  the  forces  ; 
but  when  these  points  are  given  the  equation  of  moments 
is  not  only  simplified,  but  it  is  found  that  there  is  always 
a  point  on  the  line  of  action  of  the  resultant,  called  the 
centre  of  parallel  forces  ,  which  possesses  an  important 
property. 

200.  Centre  of  Parallel  Foroes.  —  The  centre  of  par- 
allel forces  is  that  point  through  which  the  resultant  will 
constantly  pass  as  the  forces  are  rotated  about  their  points 
of  application,  the  forces  remaining  constantly  parallel  as 
they  are  rotated. 

To  illustrate,  let  the  parallel  forces  P,  F,  and  R  be  in 
equilibrium,  having  their  points  of  application  at  P,  A, 
and  G  in  the  straight  line  PC. 
The  point  C  will  be  the  point  of 
application  of  the  resultant  of  P 
and  JR.  Draw  CE  perpendicular 
to  the  lines  of  action  of  the  forces  ; 

then,  since  there  is   equilibrium, 
we 


P.EC=  R.DG.   .        .        .    (1) 

Conceive  that  the  forces  are  revolved  through  an  angle 
of  90°,  retaining  their  relative  directions  of  action.  The 
force  R  will  then  act  at  A,  parallel  to  .EC  and  to  the 
right  (or  left),  and  P  will  act  at  the  point  P,  also  parallel 


[201.]  PARALLEL    FORCES.  129 

to  EC  and  to  the  left  (or  right).  In  the  new  position  the 
arm  of  R,  in  reference  to  (7,  will  be  DA,  and  of  P,  will 
be  EP  ;  hence,  if  there  is  equilibrium,  we  will  have 

P.EP  =  R.DA.    .        .        .    (2) 
For,  from  the  similar  triangles  CD  A  and  CEP,  we  have 

EC:DC\:EP-.DA. 


which,  substituted  in  equation  (1),  gives 
P.EP^R.DA-, 

which  is  the  same  as  equation  (2)  ;  hence,  the  resultant  in 
the  new  position  will  pass  through  C. 

In  a  similar  way  it  may  be  shown  that  it  will  pass 
through  C  for  any  amount  of  rotation  of  the  forces  P  and 
R;  hence,  C  is  the  centre  of  the  parallel  forces  P 
and  J?. 

Similarly,  A  is  the  centre  of  the  parallel  forces  F  and 
P,  if  C  and  P  are  the  points  of  application  of  the  re- 
spective forces.  But  if  E  be  the  point  of  application  of 
P,  and  C  of  F,  then  A  will  not  be  the  centre  of  those 
forces. 

The  centre  of  two  parallel  forces  will  be  on  the  line 
joining  their  points  of  application  ;  and  by  combining 
their  resultant  with  a  third  force  in  a  similar  way,  the 
centre  of  three  forces  may  be  found,  and  so  on  for  any 
number  of  forces. 

201.  To  find  the  Centre  of  any  Number  of  Paral- 
lel Forces.—  Let  the  forces  be  in  the  plane  an/,  their 
points  of  application  being  at  A,  C,  E,  etc.  The  centre 
6* 


130 


STATICS. 


[201.] 


of  the  forces  1*1  and  fl  will  be  at  JB,  where  their  resultant 
intersects  the  line  A  C.    Let 

a?!,  y^  be  the  coordinates  of  A, 
rt>    m          «  u  n 

®Zl  2/2  ^, 

T        tl  "  "  A' 

^j  ys  -&> 


and 


be  the  coordinates  of  the  centre  of  all 
the  forces,  which,  being  on  the  result- 
ant, is  called  the  point  of  application 
of  the  resultant. 


Fio.  75. 

Draw  ./ig-  and  JSr  parallel  to  OX,  Am,  Bn,  etc.,  parallel 
to  OY,  and  Oe  perpendicular  to  the  lines  of  action  of  the 
forces.     The  triangles  AqB  and  BrC  are  similar,  and  give 
AB \BC\\Aq\Br  \ 

\\On-Om\Op--On 
n  x'  —  xl :  x^  —  a/'. 


[202.]  PARALLEL    FORCES.  131 

But,  in  reference  to  the  point  JB,  we  have 

'$£.'&  &  If.  M\ 

and  because  ac  and  .A6rare  cut  by  parallel  lines,  we  have 


which,  by  means  of  the  preceding  proportion,  gives 

db  :  bo  :  :  x'  •—  &i  :  x.z  —  x'  ; 
and  this,  in  the  above  equation,  gives 


or, 

which,  by  transposing,  gives 
(Fl+Fz}x' 

.-.  R&'  — 

In  a  similar  way  we  would  find 
and,  finally, 


etc.  =  SFx 
and  similarly, 

By  =  F&  +  F&2  +  etc.  =  2Fy. 
From  these  equations,  we  have 


_ 
tt 


202.  If  the  System  be  referred  to  Three  Rectangu- 
lar Axes,  a?,  y,  s,  in  which  a,  8,  and  7  are  the  angles  which 


132  STATICS.  [203.] 

the  lines  of  action  of  the  forces  make  with  the  respective 
axes  ;  then  the  equations  for  equilibrium  will  be 

X  =  cos  a%F=  It  cos  a  • 
T—  cos  @SF=  R  cos  ft  ; 
Z  =  cos  y2F=  R  cos  7  ; 


The  last  three  equations  are  the  moments  of  the  forces 
in  reference  to  the  respective  coordinate  planes. 

If  the  given  forces  are  in  equilibrium  in  reference  to 
translation,  we  have 

R  =  0; 

and,  if  they  are  also  in  equilibrium  in  reference  to  rota- 
tion, we  have 

SFx  =  0  ;     SFy  =  0  ;     SFs  -  0 


~d;    y~o;       =o; 

hence,  the  centre  of  an  equilibrated  system  is  indetermi- 
nate. 

203.  Centre  of  a  Mass. — The  centre  of  the  mass 
which  constitutes  a  body,  is  a  point  so  situated  that,  if  its 
distance  from  any  axis  be  multiplied  by  the  entire  mass, 
the  product  will  equal  the  sum,  of  the  products  obtained 
by  multiplying  each  elementary  mass  by  its  distance  from, 
the  same  axis. 

This  point  will  be  determined  when  its  position  in 
reference  to  three  rectangular  planes  is  known. 


[208.]  PARALLEL    FOECES. 

Let  M  =  the  total  mass  of  the  body  ; 
m  =  an  elementary  mass  ; 
x,  y,  0,  the  coordinates  of  any  element,  and 
x,  y,  z,  the  coordinates  of  the  centre  of  the  mass  ; 
then,  according  to  the  definition,  we  have 


=  Smy, 

Mz  =  2ms. 

These  equations  are  applicable  to  several  bodies.  If 
the  origin  of  coordinates  be  at  the  centre  of  the  mass,  we 
have 

5  =  0;    y  =  0;    z  —  ^ 
.:  2mx  =  0  ;    Smy  =  0  ;    Sma  =  0. 

EXAMPLES. 

1  Two  parallel  forces  whose  magnitudes  are  6  and  11, 
acting  in  the  same  direction  upon  a  rigid  line,  have 
their  points  of  application,  A  and  £,  5  feet  from 
each  other  ;  required  the  point  of  application  of  the 
resultant. 

2.  In  the  preceding  example,  find  the  point  of  application 

of  the  resultant  if  the  forces  act  in  contrary  direc- 
tions. 

3.  If  the  weights  2,  3,  4,  and  5  Ibs.  act  perpendicularly  to 

a  straight  line  at  the  respective  distances  of  2,  3,  4, 
and  5  feet  from  one  extremity,  what  will  be  their 
resultant  and  its  point  of  application? 

4.  Let  the  weights  3,  4,  5,  and  6  act  perpendicularly  to  a 

straight  line  at  the  points  A,  J3,  G,  and  D,  the  dis- 
tances AB  =  3  feet,  BO  =  4  feet,  and  AD  —  5  feet  ; 
required  the  resultant,  and  the  distance  from  A  to 
the  point  of  application  2?  of  the  resultant. 


134  STATICS. 

5.  If  two  parallel  forces,  P  and  F,  act  in  contrary  direc- 
tions at  the  points  A  and  B,  and  make  an  angle  $ 
with  the  line  AS ;  find  the  moment  of  each  in  refer- 
ence to  the  point  of  application  of  the  resultant. 

EXERCISES. 

1.  Has  a  statical  couple  a  centre  of  force  ? 

2.  Will  several  parallel  forces  always  be  in  equilibrium  if  the  sum  of 

their  moments  is  zero  ? 

3.  When  may  the  resultant  of  parallel  forces  be  zero,  and  the  system  not 

be  in  equilibrium  ? 

4.  If  a  system  is  in  equilibrium  why  may  the  centre  of  force  be  at  any 

point  ? 

5.  If  the  mass  of  a  body  is  homogeneous,  will  the  centre  of  the  mass  be 

at  the  geometrical  centre  of  the  body  ? 

6.  If ,  in  a  sphere,  the  density  varies  directly  as  the  distance  from  the 

centre  in  all  directions,  will  the  centre  of  the  mass  be  at  the  centre 
of  the  sphere  ? 

7.  State  different  laws  according  to  which  the  density  in  a  sphere  may 

vary,  and  have  the  centre  of  the  mass  at  the  centre  of  the  sphere. 


.       CHAPTER  IX. 

CENTRE    OF    GRAVITY. 

204.  The  lines  of  action  of  the  force  of  gravity  con- 
verge towards  the  centre  of  the  earth ;  but  the  distance 
of  the  centre  of  the  earth  from  the  bodies  which  we  have 
occasion  to  consider,  compared  with  the  size   of   those 
bodies,  is  so  great,  that  we  may  consider  the  lines  of  action 
of  the  forces  as  parallel.     The  number  of  the  forces  of 
gravity  acting  upon  a  body  may  be  considered  as  equal  to 
the  number  of  particles  composing  the  body. 

205.  The  Centre  of  Gravity  of  a  bodym&y  be  defined 
as  the  centre  of  the  parallel  forces  of  gravity  acting  upon 
the  body  ;  and  hence  the  centre  of  gravity  of  bodies  may 
be  found  in  the  same  way  as  the  centre  of  parallel  forces. 

206.  The  Resultant  of  the  Force  of  Gravity  equals 
the  weight  of  the  body  ;  or 

£  =  W. 

207.  If  a  body  be  supported  at  its  centre  of  gravity,  and 
the  body  be  turned  about  that  point,  it  will  remain  in 
equilibrium  in  all  positions,  for  it  will  be  equivalent  to 
turning  the  forces  through  the  same  angle. 

208.  Proposition. — If  a  body  be   suspended  at   any 
point,  then,  for  equilibrium,  the  vertical  through  the  cen- 
tre of  gravity  will  pass  through  the  point  of  support. 

Let  the  body  be  suspended  at  the  point  c,  the  centre  of 
gravity  being  at  a.  If  the  vertical  through  a  does  not 
pass  through  c,  join  the  points  c  and  a  by  the  line  ca}  draw 


136 


STATICS. 


[209,  210.  \ 


the  vertical  cb,  and  the  horizontal  ab.  The  weight  W  may 
be  represented  by  the  line  cb,  of  wliich  the  components 
are  ba  and  ca.  The  component  ba  will  cause  the  body  to 
turn  about  c,  causing  the  centre  a  to  approach  the  vertical 
cb.  If  a  be  in  the  vertical  cb,  either  above  or  below  the 
support,  there  will  be  no  horizontal  component,  and  the 
body  wi)l  be  in  equilibrium. 


-I 


Fia.  78. 

209.  Stable  Equilibrium. — In  stable   equilibrium,   if 
the  body  be  turned  slightly  from  its  position  of  rest,  it 
will  tend  to  return  to  its  former  position.     Thus,  the  body 
represented  in  Fig.  77  is  in  stable  equilibrium.    It  appears 
also  from  Figs.  76  and  77,  that  when  the  equilibrium  is 
stable,  and  the  body  is  turned  about  the  support,  the  centre 
of  gravity  will  be  raised.     In  this  case,  therefore,  the  cen- 
tre of  gravity  is  the  lowest  possible.     The  measure  of  the 
stability  is  the  amount  which  the  centre  of  gravity  is  raised 
in  overturning  the  body. 

210.  Unstable  Equilibrium. — A  body  is  in  a  condition 
of  unstable  equilibrium,  if,  when  it  is  turned  slightly  from 
its  position  of  rest,  it  departs,  or  tends  to  depart,  farther 
from  that  position.     This  is  illustrated  by  Fig.  78.     In 
this  case  the  centre  of  gravity  will  fall,  when  the  body  ia 
turned  about  its  support.     The  centre  of  gravity  will  be 
the  highest  possible. 


CENTRE    OF    GEAVITY. 


137 


211.  Indifferent  Equilibrium  is  that  condition,  in  which 
a  body  will  remain  in  equilibrium  after  being  slightly  dis- 
turbed.    A  sphere  or  cone  resting  on  a  horizontal  plane 
is  an  example  of  indifferent,  or  neutral,  equilibrium. 

212.  Trial  Methods. — The  preceding  principles  enable 
one  to  determine,  in  an  experimental  manner,  the  centre 
of  gravity  of  bodies.     Thus, 

in  Fig.  79,  let  the  body  be 
carefully  balanced  upon  a 
knife-edge,  and,  when  bal- 
anced, the  line  of  support  be 
carefully  marked  upon  the 
body.  Then  balance  it  in  a 
similar  way  along  another 

line ;  the  intersection  of  the  lines  will  be  vertically  under 
the  centre  of  gravity,  and,  if  the  body  be  a  thin  plate,  we 
say  that  the  centre  of  gravity  is  at  their  intersection. 

213.  If  a  body  be  suspended  at  a  point  a,  the  centre  of 
gravity  will  be  vertically  under  it.     Draw  the  vertical 
line  ab   upon   (or  within)  the  body;  then  suspend  it  at 


L 


FIQ.  79. 


Fio.   50. 


Fio.  81. 


another  point  c,  Fig.  81,  and  mark  the  vertical  line  cd. 
The  centre  of  gravity  will  be  at  g,  the  intersection  of  the 
lines  ab  and  cd. 


138  STATICS.  [213, 214.J 


EXERCISES. 

1.  If  a  carriage  stands  upon  a  side  hill,  what  condition  must  be  fulfilled 

in  order  that  it  shall  not  overturn  ?    What  must  be  the  condition 
that  it  shall  overturn  ? 

2.  A  man  stands  upon  a  floor  ;  how  far  can  he  lean  forward  or  backward 

and  not  fall  over  ? 

3.  When  a  man  moves  his  head  forward,  what  other  motion  must  his 

body  have  that  he  may  remain  in  equilibrium  upon  his  feet  ? 

4.  Why  will  not  a  table  be  as  stable  when  standing  upon  two  legs  as 

upon  three  ? 

5.  Why  is  it  more  difficult  to  overturn  a  body  like  Fig.  77  than  it  is 

one  like  Fig.  78,  the  bodies  being  of  equal  weight  ?. 

6.  If  a  book  be  suspended  at  one  corner,  why  will  its  edges  be  inclined 

to  a  vertical  ? 

7.  May  a  body  be  in  a  state  of  neutral  equilibrium  in  reference  to  a 

disturbance  in  one  direction,  and  stable  in  reference  to  another  ? 


FIG.  82.  FIG.  83. 


8.  Explain  how  the  toy  shown  in  Fig.  82  may  be  in  stable  equilibrium. 

9.  Explain  how  the  toy  horse  shown  in  Fig.  83  stands  upon  the  post 

without  falling  off. 

Centre  of  Gravity  of  Heavy  Particles. 

214.  Centre  of  Gravity  of  two   Particles. — Let  P 
be  the  weight  of  a  particle  at  A,  and 

O TC o      W,  that  at  C.     The  centre  of  gravity 

will  be  at  some  point  j5,  on  the  line  join- 
ing A  and  C.    Hence, 


r  215,  216.] 


CENTRE    OF    GRAVITY. 


139 


but, 

AJB  +  BC=AC-y 

which,  combined  with  the  preceding  equation,  gives 

W 


If 
then 


215.  Centre  of  Gravity  of  several  Heavy  Parti- 
cles.— Let  WH  w2,  ws,  etc.,  be  the  weights  of  the  particles. 
Join  wv  and  ws  by  a  straight  line  and  find  their  centre  of 
gravity  A,  as  in  the  preceding  article.  Join  A  with  w2 
and  find  the  centre  of  gravity  J?,  which  will  be  the  centre 
of  gravity  of  the  three  weights  w3,  w^  wz.  In  a  similar 
way  find  C,  the  centre  of  gravity  of  the  four  weights.  In 
this  way  the  centre  of  gravity  of  any  number  of  weights 


FIG.  85. 


FIG.  86. 


may  be  found.  It  is  not  necessary  that  the  weights  be  in 
one  plane ;  they  may  be  distributed  in  any  manner  in 
space. 

216.  If  the  Positions  of  the  Particles  are  referred 
to  Three  Rectangular  Axes,  let 


140  STATICS.  [217.] 

MI,  w2,  io3,  etc.,  be  the  weights  of   the  respective 

particles, 
xu  yi)  Zi  the  coordinates  of  wt  and  similarly  for  w^ 

ws,  etc., 
x,   yt    ~z  the  coordinates  of  the  centre  of  gravity  of 

all  the  weights,  and 
TFthe  sum  of  all  the  weights  ; 
then,  we  have 

and,  according  to  Article  202,  we  have 


EXAMPLES. 

1.  Two  particles  are  joined  by  a  straight  line  ;  if  one  is  n 

times  as  heavy  as  the  other,  find  the  position  of  the 
centre  of  gravity. 

2.  If  three  equal  particles  are  at  the  vertices  of  a  triangle, 

find  the  position  of  the  centre  of  gravity. 

3.  If  the  weights  of  three  particles  are  as  1  to  2  to  3,  and 

are  placed  at  the  vertices  of  an  equilateral  triangle, 
find  the  position  of  the  centre  of  gravity. 

4.  If  four  equal  weights  are  at  the  vertices  of  a  triangular 

pyramid,  find  the  position  of  the  centre  of  gravity. 

Centre  of  Gravity  of  Lines. 

217.  Straight  Lines. — By  a  line,  we  here  mean  a  mate- 
rial line,  whose  transverse  section  is  very  small,  such  as  a 
very  fine  wire. 

The  centre  of  gravity  of  a  uniform  straight  line  is  at 
its  middle  point. 


CENTRE    OF    GRAVITY. 


141 


For,  we  may  conceive  it  to  be  composed  of  pairs  of 
particles,  each  of  which  is  at  the  same  distance  from  the 
middle  point;  and  as  this  point  will  be  the  common  centre 
of  gravity  of  all  the  pairs,  it  will  be  the  centre  of  gravity 
of  the  line. 

218.  Centre  of  Gravity  of  the  Perimeter  of  a  Tri- 
angle. —  Let  ABC  be  the  triangle.  The  centre  of  gravity 
of  the  sides  will  be  at 
their  middle  points,  D, 
E,  F.  Join  these  points. 
The  weight  at  E  will  be 
to  that  at  D  as  the  length 
of  OB  is  to  the  length 
of  A  C.  Divide  the  line 
DE  at  the  point  G,  so 
that 

DG  \GE\\BC\  AC, 

then  the  point  G  will  be  the  centre  of  gravity  of  the  two 
lines  A  G  and  CB.  Similarly,  /  will  be  the  centre  of 
gravity  of  B  C  and  AB,  and  H,  that  of  AB  and  A  C.  The 
construction  gives 


hence,  the  preceding  proportion  gives 

DG  :  GE\\\BC  \\AC 

:  :  DF  :  FE. 

Therefore,  the  line  drawn  from  F  to  G  will  bisect  the 
angle  F\  and,  similarly,  for  DI  and  EH. 

The  centre  of  gravity  of  the  three  sides  of  the  triangle 
ABC,  will  be  in  the  line  GF;  and,  similarly,  it  will  be  in 
the  lines  Z>/and  HE\  hence,  it  will  be  at  their  intersec- 
tion, which  will  be  ike  centre  of  the  circle  inscribed  in  the 
triangle  DEF. 


142  STATICS.  [219,220.^ 

219.  Symmetrical  Lines. — The  centre  of  gravity  of 
lines  which  are  symmetrical  in  reference  to  a  point,  will 
be  at  that  point.  Thus  : — 

The  centre  of  gravity  of  the  circumference  of  a  circle, 

or  an  ellipse,  is  at  the  geometrical  centres  of  those  figures  : 

The  centre  of  gravity  of  the  perimeter  of  an  equilateral 

triangle,  or  of  a  regular  polygon,  is  at  the  centre  of  the 

inscribed  circle : 

The  centre  of  gravity  of  the  perimeter  of  a  square,  rec- 
tangle or  parallelogram,  is  at  the  intersection  of  the  diag- 
onals of  those  figures. 

220.  Centre  of  Gravity  of  a 
Circular  Arc. — Let  ABC  be  an 
arc  of  a  circle,  O  its  centre,  B  its 
middle  point,  and  AC  its  chord. 
The  centre  of  gravity  will  be  on 
the  radius  BO,  at  some  point  c, 
such  that  (see  Article  234), 

Arc  ABC :  radius  BO  : :  chord  A  C :  Oc. 
BO.  AC 
-ABO-' 

EXAMPLES. 

1.  Find  the  position  of  the  centre  of  gravity  of  the  edges 

of  a  rectangular  box. 

2.  Find  the  position  of  the  centre  of  gravity  of  the  edges 

of  a  regular  pyramid  having  a  square  base,  and 
whose  altitude  equals  the  length  of  one  side  of  the 
base. 

3.  Find  the  centre  of  gravity  of  the  semi-circumference 

of  a  circle.  .  %r 

Ana.  Oc  =  — . 


[221,222.]  CENTRE    OF    GRAVITY  143 

4.  In  Fig.  88,  if  the  angle  AOC  =  60°,  find  the  position 

of  the  centre  of  gravity  of  the  arc  AS  C. 

3r 
Ans.  Oc=  — . 

7T 

5.  Find  the  distance  from  the  centre  of  a  circle  to  the 

centre  of  gravity  of  a  quarter  of  the  circumference 
of  the  circle. 

Ans.  Oc  =  2V%-- 

7T 

Centre  of  Gravity  of  Surfaces. 

9.9.1.  Definition. — A  surface  here  means  a  very  thin 
plate  or  shell. 
222.  The  Centre  of  Gravity  of  a  Plane  Triangle 

is  in  the  line  joining  the  vertex  with  the  middle  point  of 
the  base,  and  at  one-third  the  length  of  the  line  from  the 
base. 

Let  ABO  be  a  triangle.     Consider  the  triangle  as  com- 
posed of  an  indefinitely  large  number  of  straight  lines 
parallel  to  the  base  AB.     The  centre 
of  gravity  of  each  of  these  lines  is  at 
their  middle  point ;  hence,  the  centre 
of  all  of  them  will  be  at  some  point  in 
the  line  passing  through  their  centres; 
which  point  will  be  the  centre  of  grav- 
ity of  the  triangle.     Let  D  be  the  mid- 
dle point  of  AS ;  join  C  and  D,  then 
will  the  centre  of  gravity  of  the  tri- 
angle be  in  the  line  CD.     Similarly,  it     A        ^  89 
will  be  on  the  line  AE,  drawn  from  A 
to  the  middle  point  of  BC\  hence,  it  will  be  at  g,  the  inter- 
section of  DC  and  AE.     To  find  the  distance  Cg,  draw 
DE,  then  will  the  similar  triangles  DEg  and  AgC  give 


144  STATICS.  |223,  2SJ4.J 


AC 
.:g 
Adding  g*D  to  both  members,  gives 

Similarly, 


We  may  readily  find  that,  the  perpendicular  distance  of 
the  centre  of  gravity  from  the  base,  equals  one-third  of 
the  altitude. 

223.  Symmetrical  Figures.  —  The  centre  of  gravity  of 
the  surface  of  a  circle,  or  of  an  ellipse,  is  at  the  geomet- 
rical centre  of  the  figure  ;  of  an  equilateral  triangle,  or  a 
regular  polygon,  it  is  at  the  centre  of  the  inscribed  circle  ; 
of  a  parallelogram,  at  the  intersection  of  the  diagonals  ; 
of  the  surface  of  a  sphere,  or  an  ellipsoid  of  revolution,  at 
the  geometrical  centre  of  the  body  ;  of  the  convex  sur- 
face of  a  right  cylinder,  at  the  middle  point  of  the  axis 
of  the  cylinder.  The  centre  of  gravity  of  the  convex  sur- 

face of  a  regular  right  pyramid, 
or  of  a  right  cone  having  a  circle 
for  its  base,  is  on  the  axis  of  the 
figure  at  one-third  the  altitude 
from  the  base  ;  for,  the  surface 
may  be  considered  as  composed  of 
triangles  having  a  common  apex. 
224.  To  find  the  Centre  of 

FIG.  90. 

Gravity  of  a  Part  of  a  Body, 

when  the  centre  of  gravity  of  the  whole  and  of  the  remain- 
tngpart  are  known. 

Let  AB  be  one  area,  CD  the  other,  O  the  centre  of 


[225,226.]  CENTRE    OP    GRAVITY.  145 

gravity  of  the  former,  o  that  of  the  latter,  and  c  the  centre 
of  gravity  of  the  part  remaining  after  removing  the  area 
CD.  The  point  c  will  be  on  the  line  through  oO.  Take 
A  as  the  origin  of  moments.  The  weights  are  directly 
proportional  to  the  areas,  and  the  moment  of  the  whole 
equals  the- sum  of  the  moments  of  all  the  parts,  hence 

Area  AB  x  A  O=Area  CD  x  Ao  -f  (Area  AB—Area  CD) 
Area  AB  xAO  —  Area  CD  x  Ao 

•      ju  /»  ^— • •  _     -  -  - -    - 

Area  AB— Area  CD 

The  same  formula  will  apply  to  lines  and  volumes  by 
simply  substituting  line  or  volume  for  area. 

225.  Irregular  Figures. — Any  figure  may  be  divided 
into  rectangles  and  triangles,  and,  the  centre  of  gravity  of 
each  being  found,  the  centre  of  gravity  of  the  whole  may 
be  determined  by  treating  it  as  if  it  were  an  aggregation  of 
particles,  as  in  Articles  215  and  216. 

226.  The  Centre  of  Gravity  of  a  Zone  is  at  the  mid- 
dle point  of  the  line  joining  the  centres  of  the  upper  and 
lower  bases  of  the  zone. 

It  is  proved  in  Geometry  that,  on  the  same  or  equal 
spheres,  zones  are  to  each  other  as  their  altitudes  ;  hence, 
if  the  zone  be  divided  into  an  indefinite  number  of  paral- 
lel zones,  and  all  be  reduced,  or  contracted,  to  the  axis  of 
the  zone,  it  will  form  a  line  of  uniform  weight ;  and 
hence,  the  centre  of  gravity  will  be  at  the  middle  point  of 
the  line. 

EXAMPLES. 

1.  If  a  line  be  drawn  parallel  to  the  base  of  a  triangle, 
dividing  it  into  equal  areas,  will  it  pass  through  the 
centre  of  gravity  of  the  triangle  ? 

7 


146 


STATICS. 


[226.] 


2.  If  a  line  bisects  the  vertical  angle  of  a  triangle,  in  what 

cases  will  it  pass  through  the  centre  of  gravity,  arid 
in  what  cases  will  it  not  ? 

3.  If  the  bases  of  two  triangles  are  in  the  same  line,  and 

their  vertices  are  in  a  line  parallel  to  the  bases,  show 
that  the  line  joining  their  centres  of  gravity  will  also 
be  parallel  to  the  bases. 

4.  In  Fig.  90,  if  the  circles  are  tangent  to  each  other  in- 

ternally at  A,  find  the  distance  from  A  to  the  centre 
of  gravity  c,  after  the  smaller  circle  has  been  re- 
moved. Let  R  =  AO ;  r  =  Co. 

Ans.  Ac  =  -F» — 5. 


5.  Find  the  centre  of  gravity  of  the  remainder  of  a  square 
after  one- quarter  of  it  has  been  removed  from  one 
corner. 

Ans.  AB  = 


D/£L 


Fio.  91. 


V 


F 
FIG.  92. 


- 


6.  Find  the  centre  of  gravity  of  a  trapezoid. 

It  will  be  on  the  line  joining  the  centres  of  the  twc 
bases. 


F227.] 


CENTRE    OP    GRAVITY. 


147 


7.  Find  the  centre  of  gravity  of  the  surface  of  a  right 
cone  having  a  circular  base,  including  the  base. 
Let  r  be  the  radius  of  the  base  and  h  the  altitude  ; 
find  the  distance  from  the  apex. 


Am. 


7 
h. 


Centre  of  Gravity  of  Volumes. 

227.  Triangular  Pyramid.  —  The  centre  of  gravity  of 
any  triangular  pyramid  is  on  the  line  joining  any 
apex  with  the  centre  of  gravity  of  the  opposite  face,  and  at 
a  point  three-fourths  the  length  of  the  line  from  the  apex. 

Let  A-BCD  be  a 
triangular  pyramid. 
Suppose  that  it  is  di- 
vided into  infinitely 
thin  slices,  bed,  parallel 
to  the  base,  BCD.  The 
centre  of  gravity  of  the 
pyramid  will  be  on  the 
line  passing  through 
the  centres  of  all  the 
slices.  Let  F  be  the 
centre  of  gravity  of  the 
base,  then  will  the  cen- 
tre of  gravity  of  the  pyramid  be  on  the  line  AF.  Simi- 
larly, it  will  be  on  the  line  BG  drawn  from  the  apex  B  to 
the  centre  of  gravity  of  the  opposite  face,  and  hence,  at 
their  intersection  H.  Join  ^and  6r,  and  the  similar  tri- 
angles jf^/Lfi'and  AEB  give 

EG  _    FG_ 

•  AE 


FIG.  93. 


148 


STATICS. 


[228,229. 


But,  EG  =  \AE\ 

.-.  Fa  =  \AB. 

The  similar  triangles,  FGIT  and  AHJB,  give 
FG_      FIT 
AB  ~  AH> 
in  which,  substitute  the  value  of  FG,  and  it  gives 

3FH=  AH. 
Add  FH=  Ftf, 

and  we  have  4:Fff=  AF\ 

.-.  FH=  \AF. 

228.  The  Centre  of  Gravity  of  any  Pyramid  01 
Cone  is  on  the  line  joining  the  apex  with  the  centre  of 
gravity  of  the  base  and  at  one-fourth  the  distance  from 
the  base. 

That  it  will  be  on  this  line  is  evident  from  the  preced- 
ing Article.  The  pyramid  may  be  divided  into  triangular 
pyramids,  and  the  centre  of  each  will  be  in  a  plane  p.aral- 
lel  to  the  base  and  at  one-quarter  the  altitude  from  the 
base  ;  hence,  it  will  be  at  the  point  where  this  line  inter- 
sects the  plane.  The  position  for  the  cone  is  found  in  the 

same  way,  for,  the  cone  may  be 
considered  as  composed  of  an  in- 
definite number  of  pyramids. 

229.  Problem.  —  Find  the 
centre  of  gravity  of  a  spherical 
sector  generated  by  the  revolu- 
tion of  the  circular  sector  AGO 
about  the  axis  GC. 

It  will  be  on  the  axis  G  C.  If 
we  consider  that  the  spherical 
sector  is  composed  of  an  indefinite  number  of  cones,  having 
their  bases  in  the  surface  of  the  sphere,  and  their  apices  at 


[230.]  CENTRE    OF    GRAVITY. 

the  centre  C  of  the  sphere,  the  locus  of  the  centre  of  gravity 
of  all  the  cones  will  be  in  a  spherical  surface  DEF^  having 
C  for  a  centre  and  radius  CD  =  £  of  CA.  The  centre  of 
gravity  g  of  this  surface  will  be  the  centre  of  gravity  of 
the  spherical  sector  ;  but,  according  to  Article  226,  Eg  is 
one-half  of  the  altitude  EK  of  the  spherical  surface 
DEF.  But,  from  the  figure,  we  have 


.'.  Eg  = 
and 

Gg  =  GE+Eg 


But, 

Cg  =  CG-Gg 

=  CG  -\CG-\GH 
=  f(2<7#-  GH). 

230.  Problem.  —  To  find  the  centre  of  gravity  of  a  seg- 
ment of  a  sphere. 

Let  AGJBj  Fig.  94,  be  the  segment  of  a  sphere,  and  a 
point  g'  on  the  line  GH  be  its  centre  of  gravity.  Taking 
the  origin  of  moments  at  the  centre  C,  we  have 

Vol.  ofSeg.  x  Cg'  =  Vol:  of  sector  AGBCx  Cg 
—  Vol.  of  cone 


or, 

2(  CG-IGH]  .  Cg'  =  ^(CGf.        .\GC  .  Cg 
.  \CH.ICH  \ 


„  ,  _  %(CG?.GH.Cg  - 


150  STATICS.  [231,232.1 

EXAMPLES. 

1.  Find  the  centre  of  gravity  of  a  hemisphere . 

It  will  be  on  the  radius  perpendicular  to  the  base 
of  the  hemisphere,  and,  according  to  Article  229, 
at  a  distance  f  the  radius  from  the  centre. 

2.  Find  the  centre  of  gravity  of  the  remainder  of  a  sphere 

whose  radius  is  R,  after  another  sphere,  whose  radius 
is  r,  is  taken  from  it,  the  two  spheres  having  a  com- 
mon tangent  plane. 

3.  Find  the  distance  from  the  centre  of  a  sphere  to  the 

centre  of  gravity  of  a  segment  of  the  sphere  of  one 

base,  the  chord  of  the  segment  being  \  the  radius. 
4c.  A  cone  is  suspended  at  a  point  in  the  circumference  of 

the  base ;  required  the  inclination  6  of  the  axis  to 

the  horizontal.     Let  the  radius  of  the  base  be   2 

inches  and  the  altitude  8  inches. 
5.  In  the  preceding  example,  what  will  be  the  relation 

between  the  radius  of  the  base  and  altitude  of  the 

pyramid,  if  the  axis  is  inclined  30°  ? 

Centrobarid  Method. 

231.  The  two  following  theorems  are  by  some  accredited 
to  Guldinus  and  by  others  to  Pappus,  one  or  the  other  of 
whom  is  supposed  to  have  discovered  them. 

232.  Theorem  I. — The  surface,  generated  by  the  revo- 
lution of  a  line  about  an  axis  fixed  in  the  plane  of  the  line, 
is  equivalent  to  the  product  of  the  length  of  the  line  into 
the  circumference  passed  over  by  the  centre  of  gravity  of 
the  line. 

Let  AN  be  a  plane  curve,  and  YY  the  fixed  axis  in 
the  plane  of  the   curve.     Draw  any  number  of  equal 


r232.'-  CENTROBARIC    METHOD.  151 

chords,  AB,  -BO,  etc.,  and  from  their  middle  points,  a^,  a^ 

etc.,  draw  the  perpendiculars  a^b^  aj)^,  etc.,  to  the  axis 

YT.      The   revolution   of 

the  curve  about  the  axis  will 

generate   a  double  curved 

surface,  and  the   polygon, 

several  f  rustra  of  cones  in- 

scribed within  the  former 

surface. 

The  surface  generated  by 
AB  will  be  (see  Geometry) 
x  AB. 


Similarly,  the  entire  sur- 
face generated  by  the  polygon  will  be 

STT^A  x  A  B  +  #A  x  -B  U+  etc-)- 

Let  g  be  the  centre  of  gravity  of  all  the  lines  AB,  BC, 
etc.,  and  go,  a  perpendicular  to  YY,  then  will  the  mo- 
ments of  the  lines  in  reference  to  the  axis  YY  be 
gc(AB  +  B  O+  etc.)  =  AB.  a  A  +  B  C.  aj>%  -f  etc. 

Multiplying  both  members  of  this  equation  by  27r,  gives 
%irgc(AB  +  £C+  etc.)  =  &ir(AB.  «A  -f  B  C.  aj>2  +  etc.), 
the  second  member  of  which  is  the  surface  generated; 
hence, 

%Trgc  x  perimeter  of  the  polygon  =  surface  generated, 
in  which  ^irgc  is  the  circumference  described  by  the  cen- 
tre of  gravity  of  the  perimeter. 

Inscribe  in  the  curve  another  polygon  of  double  the 
number  of  sides,  and  so  on  indefinitely  ;  the  limit  of  the 
polygons  is  the  arc,  and  the  limit  of  the  surface  is  the 
double  curved  surface  ;  but,  the  preceding  equation  is 
true  for  any  number  of  sides,  and  hence,  will  be  true  of 


152 


STATICS. 


[233.] 


the  limit.  Let  gl  be  the  centre  of  gravity  of  the  are,  and 
ffidi  the  perpendicular  upon  the  axis ;  then  the  equation 
becomes 

^TT^iCi  X  length  of  arc  =  surface  generated  by  the  arc. 
The  theorem  is  evidently  true  for  a  single  line,  or  for 
several  lines  of  unequal  length.     Q.  E.  D. 

233.  Theorem  n. — The  volume,  generated  by  the  revo- 
lution of  a  plane  area  about  a  fixed  axis  in  its  plane,  is 
equivalent  to  a  prism  whose  base  is 
the  area  revolved,  and  altitude,  the 
length  of  the  circumference  passed 
over  by  the  centre  of  gravity  of  the 
area.  The  plane  area  must  lie  wholly 
on  one  side  of  the  axis. 

Let  LAB . . .  FQ  be  a  plane  area, 
YY  the  fixed  axis  in  the  plane  of 
the  area.  Divide  LQ  into  equal 
parts,  LM,  MN,  etc.,  and  draw  the 
ordinates  LA,  MB,  etc.,  perpendicu- 
lar to  YY,  and  from  A,  B,  etc., 
draw  parallels  to  YY,  forming  rectangles,  as  shown  in 
the  figure.  When  the  figure  revolves  the  rectangles  will 
generate  cylinders,  and  the  curve,  a  double  curved  surface. 
The  volume  of  the  cylinders  will  be 

IT  AD.  LM+TrBM\  MAT+etc. 

Let  g  be  the  centre  of  gravity  of  all  the  rectangles,  and 
the  ordinate  to  the  centre  of  gravity  of  each  being  equal 
to  one-half  the  length  of  the  side  of  the  rectangle,  we 
have  for  the  moments  of  the  rectangles  in  reference  to  the 
axis  YY, 

gc  x  area  of  all  the  rectangles  =  AL .  LM . 


Fio.  96. 


[234.J  APPLICATIONS.  153 

/.  I2trffc  x  area  of  all  the  rectangles  =  irAL* .  LM  + 


=  Vol.  of  all  the  cylinders. 

If  now  the  divisions  in  LQ  be  increased  indefinitely, 
the  limit  of  the  rectangles  will  be  the  area  of  the  curve,  and 
the  limit  of  the  cylinders  will  be  the  volume  of  revolution. 
Let  ffi  be  the  centre  of  the  plane  area,  then  we  have 

Vol.  of  revolution  =  2<7r<71c1 .  area    of  the  plane 

curve, 

=  the  area  of  the  curve  x  by 
the  distance  described  l/y 
the  centre  of  gravity  of 
the  area.  Q.  E.  D. 

Applications. 

234.  Problem. — Tojmd  the  centre  of  gravity  of  a  cir- 
cular arc. 

Let  ABC  be  a  circular  arc  whose  centre  is  0 ;  the 
centre  of  gravity  will  be  at  some  point  c  on  the  radius  OB 
drawn  to  the  middle  point,  B,  of  the  arc. 
Through  0draw  the  axis  ZJTparallel  to  the  I 

chord  AC,  and  conceive  the   curve   to  be        / 
revolved  about  this  axis,  generating  a  zone.     •* 
The  area  of  the  zone  will  be  (see  Geometry)          V 


-D- 


y 

FIG.  97. 

According  to  Theorem  1,  we  have 

Arc  ABC.  27r6>c  =  Area  of  the  zone  =  ^OB.AC\ 

AC.  OB 


154- 


STATICS. 


[285.] 


235.  Problem. — Find  the  distance  from  the  centre  of  a 
circle  to  the  centre  of  gravity  of  a  sector  of  the  cirde. 

Let  AGBC  be  the  sector  generated  by  the  rotation  of 
the  line  A  C  about  the  centre  C.     Let  6  be  the  angle  A  CG, 

and  r  the  radius  AC;  then  will 
the  arc  AGB=  ZrQ.  Accord-' 
ing  to  Theorem  Jt'we  find  that 
the  area  of  the  sector  will  be  ra#. 
If  the  sector  be  divided  into 
an  indefinitely  large  number  of 
sectors,  each  may  be  considered 
as  a  triangle  whose  centre  of 
gravity  is  at  two-thirds  of  its  alti- 
tude from  the  centre  C.  With 
a  radius  DC,  equal  to  fr,  describe  the  arc  DEF ';  this 
arc  will  be  the  locus  of  the  centre  of  gravity  of  all  the 
small  sectors  ;  and  the  centre  of  gravity  of  all  of  them,  or 
the  sector  A  CB  will  be  at  g,  the  centre  of  gravity  of  the 
arc  DEF.  According  to  Article  234  we  have 
2  x  I?'  sin  6 .  f  r  <.r  sin  6 


EXAMPLES. 

T 

1.  If  ACBG  is  a  semi-circle,  prove  that  Cg  is  f  -. 

2.  If  g',  Fig.  98,  be  the  centre  of  gravity  of  the  circular 

segment  A  GB,  find  the  distance  Cg1. 

3.  Find  the  volume  of  a  sphere. 

4.  Find  the  volume  generated  by  the  revolution  of  the 

circular  segment  A  GB  about  an  axis  through  C  and 
parallel  to  the  chord  AB. 

5.  Find  the  volume  generated  by  the  circular  sector  CA  GB^ 

about  an  axis  through  G  and  parallel  to  AB. 


CHAPTER  X. 

SOLUTION   OF   PROBLEMS    A.CCORDING    TO   THE   PRINCIPLES   OP 
ENERGY. 

Problems  in  which  the  Solutions  depend  upon 
Potential  Energy. 

236.  Energy  represented  by  the  Three  States  of 
Equilibrium. — According  to  Article  210,  when  a  body  ie 
in  the  condition  of  unstable  equilibrium,  the  centre  of 
gravity  is  in  the  highest  position.     In  this  condition  its 
potential  energy  is  a  maximum,  that  is,  it  is  in  a  condition 
to  do  the  most  work.     When  the  centre  of  gravity  is 
lowest,  the  body  is  in  a  condition  of  stable  equilibrium 
(Article  209),  and  its  potential  energy  is  a  minimum,  that 
is,  it  is  in  a  condition  to  do  the  least  work.     In  neutral 
equilibrium,  the  potential  energy,  for  successive  positions 
of  the  body,  remains  constant. 

237.  To  find  a  curve  such  that  a  heavy  bar  A.  E,  rest- 
ing against  it  and  against  a  vertical 

plane  DE,  will  be  in  equilibrium  in 
all  positions,  there  being  no  friction 
on  the  surfaces. 

Let  g  be  the*  centre  of  gravit}T  of 
the  bar.  If  the  bar  is  prismatic  and 
homogeneous,  g  will  be  at  the  middle 
of  the  length,  but,  in  other  cases,  it 
may  be  at  any  other  point  along  the  bar. 

This  is  a  case  of  indifferent  equilibrium,  and  hence,  the 
centre  of  gravity  is  neither  raised  nor  lowered  by  a  change 


156 


SOLUTION    OF    PROBLEMS. 


[  238.) 


of  position  ;  in  other  words,  its  locus  will  be  in  a  horizon- 
tal line.  Therefore,  if  the  centre  of  gravity  g  be  moved 
along  the  line  CJB,  and  the  end  E  be  kept  constantly 
against  the  vertical  DE,  the  end  A  will  trace  the  curve. 
But  this  is  equivalent  to  the  well-known  method  of  con- 
structing an  ellipse  by  means  of  a  trammel.  Hence,  the 
curve  CAD  is  an  ellipse. 

[Let 


then, 

and 

or, 


which  is  the  equation  of  the  ellipse.     A  more  general  solution 
will  be  found  by  letting  Ag  =  n .  Eg.} 

238.  Required  the  form  of  a  curve  such  that  a  heavy 
bar  resting  against  it  and  against  a  smooth  pin  above  the 
curve,  will  be  in  equilibrium  in  all  positions. 

Let  AD  be  the  bar,  D  the 
position  of  the  pin,  and  AB 0 
the  required  curve. 

This  is  also  a  case  in  which 
the  potential  energy  is  con- 
stant /  hence,  the  centre  of 
gravity  will  be  found  in  a 
horizontal  line  gg',  passing 
through  the  centre  of  grav- 
ity, <7,  of  the  bar  in  the  vertical  position.  The  curve  may 
therefore  be  constructed  by  drawing  any  number  of  radial 
lines  DJ3)  DA,  etc.,  through  D,  intersecting  them  by  the 
horizontal  line  gg ',  and  laying  off  on  the  radial  Hues  below 


[239,  240.]  POTENTIAL    ENERGY.  157 

the  horizontal  the  constant  distance  gB  =  g'A,  etc.     The 
curve  is  called  the  conchoid  of  Nicomodes. 
[Let 

Ag'  =  Bg  -  a ;  gD  =  s-t  ADB  =  8 ;  AD  =  p ; 
then, 

AD  =  Ag'+g'D 

or, 

p  =  a  +  c  sec  0 ; 

which  is  the  polar  equation  of  the  curve.  ] 

239.  A  cord  of  given  length  is  suspended  at  twopoinU 
in  the  same  horizontal /  required  the  form  of  the  curve 
when  the  centre  of  gravity  is  the  lowest. 

The  cord,  being  perfectly  flexible,  will  naturally  assume 
the  position  of  stable  equilibrium, 
and  its  potential  energy  will  be  a 
minimum  ;  that  is,  its  centre  of  grav- 
ity will  be  the  lowest  possible.  The 
curve  assumed  by  such  a  cord  is  called  a  Catenary.  If  the 
cord  be  of  variable  density,  it  will  still  assume  the  position 
in  which  the  centre  of  gravity  is  lowest. 

[The  equation  of  the  catenary  is  found  by  higher  mathematics. 

If  w  =  the  weight  per  unit  of  length  of  the  cord ;  t0  =  the  ten- 
sion at  the  lowest  point  of  the  cord ;  s  =  the  base  of  the  Naperian 
system  of  logarithms ;  x  horizontal  and  y  vertical ;  the  origin  of 
coordinates  being  taken  at  the  lowest  point,  then 


[WX  MKg~l  9 

6^_6~^ 


240.  A  curve  if  given  length  is  revolved  about  the  line 
passing  through  its  extremities  /  required  the  form  of 
the  curve  such  that  the  surface  generated  shall  be  a 
maximum. 


158 


SOLUTION    OF   PROBLEMS. 


[241.J 


The  area  will  equal  the  length  of  the  curve  multiplied 
by  the  distance  passed  over  by  the  centre  of  gravity  of  the 
curve.  The  curve  being  of  constant  length,  the  area 
will  therefore  be  greatest  when  the  distance  of  the  centre 
of  gravity  of  the  curve  from  the  axis  of  revolution  is 
greatest  ;  hence  the  curve  must  be  a  catenary. 

241.  Two  cylinders  of  unequal  radii,  but  of  the  same 
material  and  length,  are  placed  in  a  larger  hollow  cylinder; 
find  their  position  when  in  equilibrium. 

Their  position  will  be  that  in  which  the  potential  energy 
is  least  ;  hence,  their  centre  of  gravity  will  be  lowest. 

Let  g  be  the  centre  of 
gravity  of  the  cylinders  when 
in  contact.  If  they  be  rolled 
in  the  hollow  cylinder,  re- 
maining in  contact  with  each 
other,  the  centre  of  gravity 
will  describe  a  circle  IfgN^ 
about  the  centre  C  of  the 
hollow  cylinder,  and  the  line 
EF  joining  the  centres  of  the  cylinders  will  be  constantly 
tangent  to  this  arc  ;  hence,  when  they  are  in  the  position 
of  equilibrium,  the  line  EF  will  be  horizontal  and  the 
point  g  vertically  under  the  centre  C. 

To  find  the  angle  ECg,  let  R  =  AC,r2  =  F£,  and 
r,  =  AE. 

Then,  EF=  r^+rz  =  the  sum  of  the  radii  of  the  two 
cylinders  ;  and  the  equation  of  moments  gives 


w, 

FIG.  102. 


and  from  the  figure  we  have 


1242.J 


POTENTIAL    ENERGY 

Eg  +  gF=  rt  +  r2 


159 


and, 


/.  sin  ECg  = 


242.  Required  the  effort  necessary  to  maintain  a  body 
on  a  smooth,  inclined  plane,  the  effort  being  exerted  par- 
allel to  the  plane. 

Let  the  effort  be  exerted  by  a  body  acting  vertically,  as 
in  Fig.  103  ;  then  will  the  centre  of  gravity  of  the  two 
bodies  bo  in  the  same  horizontal  line  for  all  positions  of 
the  body  on  the  plane. 

Let  A  C  be  the  inclined  plane,  on  which  the  weight  P 
is  held  in  position  by  the  weight  TF,  all  without  friction. 
Let  a  and  d  be  the  centres  of  the 
bodies  respectively  in  one  posi- 
tion, then  will  their  centre  of 
gravity  be  at  some  point  g,  on 
the  line  joining  their  centres. 
If  the  bodies  are  moved  into  an- 
other position,  having  their  cen- 
tres respectively  at  b  and  e,  their 
centre  of  gravity  will  be  in  a 
horizontal  line  through  g. 

Since  the  energy  is  constant,  the  potential  energy  gained 
by  raising  the  weight  P  will  equal  that  lost  by  lowering 
TF,  or,  in  other  words,  the  product  of  P  into  the  vertical 
distance  through  which  it  has  been  raised,  equals  W  intc 
the  vertical  distance  through  which  it  has  been  lowered. 


A 


Fio.  103. 


H'O  SOLUTION    OF    PROBLEMS.          [243,244., 

Drawing  the  horizontal  line  be,  and  the  vertical  line  ae, 
we  have 

P.  ac  =  W.  de  ; 
but, 

ae  =  ab  sin  cba 

=  db  sin  A 
=  flfe  sin  .4 


which,  substituted  in  the  preceding  equation,  gives 

P.£C  =  W.AC-, 
or, 

P:W::AO:  BC\ 

that  is,  the  effort  is  to  the  resistance  as  the  height  of  the 
plane  is  to  its  length. 

243.  Determine  the  conditions  of  equilibrium   of  a 
single  pulley. 

In  one  position  let  the  weight  W  be  at  d,  and  P,  at  a  ; 
and  in  another  position,  P  at  b,  and  W  at  c. 
The  centre  of  gravity  will  be  at  the  same  point 
g,  in  both  positions  ;  hence,  we  have 


but, 


P      W  ab  =  dc-9 

**  1M-  .  p  -  w 

.  .  J.    -      rr  , 

that  is,  the  effort  equals  the  resistance  when  there  is  no 
friction. 

244.  Determine  the  conditions  of  equilibrium  of  the 
straight  lever. 

Let  AB  be  a  bar  in  a  horizontal  position,  having  weights 
P  and  W  suspended  at  its  extremities ;  it  is  required  tc 
find  the  point  6',  upon  which  they  will  balance.  Their 


[244.] 


POTENTIAL    ENERGY. 


161 


centre  of  gravity,  <?,  will  be  in  the  line  joining  the  centres 

of  the  bodies,  and  vertically  under  the  required  point  O. 

Let  the  arm  be  turned 

into  the   position   ab, 

the  perpendiculars  da 

and  eb  from  a  and  b 

be     dropped     upon 

AB  ;    then   will    the 

weight  W  have   been 


raised  a  height  equal 
to  be,  and  P  will  have 
fallen  a  distance  ad. 
Since  there  is  equili- 
rium,  we  have 

P.da=W.  eb. 

The  similar  triangles  ad  O  and  beO,  give 

da  _  aC  _AC 
~eb  ~' 


Fio.    106. 


and,  by  combining  these  equations,  we  find 


or 


that  is,  the  effort  is  to  the  resistance  inversely  as  the  arm 
of  the  effort  is  to  the  arm  of  the  resistance. 

In  this  problem  the  centre  of  gravity  of  the  bodies  re- 
mains at  g  for  all  inclinations  of  the  arm  AB  ;  for  the 
similar  triangles  adO  and  Ceb,  give 

dO\  Ce-.'.aC:  Ob-, 

but  dC  and  Ce  are  the  arms  of  the  forces  in  the  new  posi- 
tion ;  hence,  they  are  proportional  to  the  original  arms. 
When  the  support  is  not  in  the  line  of  the.  points  of  attach- 
ment of  the  weights,  the  centre  of  gravity  will  change  its 


SOLUTION    OF    PROBLEMS. 


[245.] 


position  as  the  arm  is  rotated,  as  will  be  seen  in  the  fol- 
lowing problem : 

245.  To  ftr<d  the  conditions  of  equilibrium  of  the  bent 
lever. 

Let  A.G  and  BG\>&  the  arms  of  the  lever,  the  support, 
or  fulcrum,  being  at  G,  and  the  weights  suspended  as 

shown  in  the  figure. 
-,.D  The  centre  of  grav- 
ity of  the  weights, 
when  in  equilibri- 
um, will  be  in  the 
line  joining  the  cen- 
tres of  P  and  W, 
and  at  a  point  g, 
vertically  under  the 
point  of  support  G. 
Let  the  lever  be 
turned  through  a  small  angle,  then  will  the  end  A  de- 
scribe the  arc  A  A',  and  £,  the  arc  SB',  and  the  weights 
will  be  found  in  the  positions  P'  and  W.  The  centre 
of  gravity  will  be  raised  to  a  point  g' ;  hence,  when  the 
bodies  are  left  to  themselves  they  will  return  to  their 
former  position. 

We  may,  however,  determine  a  practical  formula  for  this 
case  by  assuming  that  they  remain  in  equilibrium  when 
the  lever  is  turned  through  an  exceedingly  small  angle. 
For  this  case,  we  consider  the  arcs  AA'  and  BB'  as 
straight  lines.  Draw  the  horizontal  lines  A'a  and  B'b, 
then  will  the  potential  energy  of  P  be  increased  by  an 
amount  equal  to 

P.Aa; 

and  that  of  TFwill  be  diminished  by  an  amount  equal  to 


FIG.  106. 


[246.]  KINETIC    ENERGY.  163 

hence,  according  to  the  hypotheses,  we  have 
P.Aa  =  W.M. 

Through  G  draw  the  horizontal  CD,  meeting  the  verticals 
through  A  and  B  at  the  points  C  and  D.  From  the  simi- 
lar right  angled  triangles  A  A' a  and  ACG,  we  have 

Aa     _  GO 
AA'~  GA 

Similarly,  the  triangles  BB'b  and  BGD  give 

Bb    _  GD 
BB'  ~  GB ' 

But  the  arcs  BB'  and  AA  are  proportional  to  the  radii 
GB  and  G  A ;  hence, 

AA  _  BB' 

GA  '"  GB  ; 

and,  by  combining  these  three  equations  so  as  to  eliminate 
A  a  and  Bf>,  we  find 

P.GC=W.GD; 

or, 

P:W::  GDiGC; 

that  is,  the  weights  are  inversely  as  their  arms  ;  a  result 
which  agrees  with  the  preceding  problem. 

Problems  involving  Kinetic  Energy. 

246.  A  body  falls  freely  through  a  height  h  •  what  will 
be  the  kinetic  energy  stored  in  it  f 

A  body  whose  weight  is  W,  at  a  height  h  above  a  given 
point,  has  a  potential  energy  of 

Wh; 
and  when  it  has  fallen  through  this  height  its  energy  wil] 


164  SOLUTION    OF    PROBLEMS.  [247.] 

he  changed  to  kinetic  energy.     Substituting  for  A  its  value 
in  terms  of  v  (see  Eq.  (3),  Art.  72),  gives 


as  given  in  Article  112. 

If  the  body  had  fallen  through  a  portion  of  the  height 
hi,  leaving  a  height  A.,,  through  which  it  may  afterwards 
fall,  we  have  the  kinetic  energy 


and  the  potential  energy 

n 

hence,  the  total  energy  will  be 


which  is  constant  for  that  height. 

247.  Two  bodies  of  unequal  weights  are  placed  on  two 
unequally  inclined  planes,  and  connected  by  a  fine  inex- 
tensible  cord,  which  passes  over  a  pulley  so  placed  above 
the  angle  of  the  planes  that  the  cord  will  be  parallel  to  the 
planes  ;  required  the  equations  for  their  motion,  there 
being  nofrictional  resistances,  nor  resistance  of  the  air. 

Let  AC  and  BC  be  the  inclined  planes,  W  and  P  the 
positions  of  the  bodies 
when  motion  begins.  Af- 
ter a  time  t  let  them  be 
in  the  positions  represent- 
ed by  P'  and  W  ;  the  dis- 
tances over  which  they 
have  moved  being  FIG  1Q7 

dc=  ob. 
In  the  initia.  position,  the  centre  of  gravity  of  the  bodies 


1247.1  KINETIC    ENERGY.  16.") 

will  be  at  some  point  <?,  in  the  line  joining  their  respective 
centres  of  gravity  ;  and  after  a  time  t  it  will  be  at  some 
lower  position  g'.  Through  g  and  g'  draw  horizontal  lines  ; 
the  vertical  distance  ge  between  them,  will  be  the  height 
through  which  the  common  centre  of  gravity  of  both 
bodies  will  have  fallen. 
Let 

ge-h\ 

then,  the  potential  energy  lost  will  be 

(P+W}h, 
which  is  a  gain  of  kinetic  energy  equal  to 


Let 

8  =  db  —  do  ; 

then  will  the  vertical  height  through  which  P  has  fallen  be 

8  sin  B  ; 

and  the  height  through  which  W  has  been  raised  will 

be 

s  sin  A  ; 

and  the  total  potential  energy  lost  will  be 

(Psin^-TFsm^)s; 
hence,  we  have 

(P  +  W)  ^-  =  (P  &iujB  -  TFsin  A)s  ; 

y 

from  which  we  find 

,      P  sin  B  —  Wsin  A  .. 


166 


SOLUTION    OF    PROBLEMS.          [248,  249. j 


In  this  problem  the  acceleration  will  be  constant  ;  hence, 
according  to  Article  24,  we  have 


which,  substituted  in  the  preceding  and  reduced,  gives 
Psin^-TFsinJ. 


v  = 


gt.   . 


(2) 


P  +  w 

Eliminating  v  from  the  two  preceding  equations,  gives 

P  +  W 


248.  If  one  of  the  bodies  as  P,  in  the  preceding  prob- 
lem, moves  vertically,  while  the  other  moves  on  the  plane, 

required  the  formulas  for 
their  motion. 

This  problem  may  be  solved 
in  the  same  way  as  the  pre- 
ceding one;  but  the  results 
may  be  obtained  directly  from 
the  preceding  formulas  by 
making  .#  =  90°.  The  re- 
sults are 


FIG.  108. 


-SM 


'} 


P  —  W sin  A 
P  +  W 


and, 


/  r   P  +  W     ssi 

=  Y    \_P  —  WsmA'  g] 


249..  If,  in  the  preceding  problems,  both  bodies  move 
vertically,  required  the  formulas  for  the  motion. 


[250.] 


KINETIC    ENERGY. 


167 


Making  A  =  90°,  in  the  preceding  formulas,  we  find 
the  formulas  of  Problem  3,  Article  90. 

250.  A  vessel  is  filled  with  a  liquid  /  required  the  velo- 
city with  which  it  will  discharge  itself  through  an  orifice 
near  the  base. 

Let  ABE  be  the  vessel,  J^the  position  of  the  orifice. 
Suppose  that  a  small  portion,  equal  to  the  horizontal  slice 
ACB,  has  been  discharged.  The 
centre  of  gravity  of  the  mass  will 
have  been  lowered  from  some  point 
^to  another  point  <?',  and  the  poten- 
tial energy  lost  is  equal  to  the  weight 
of  the  liquid  above  the  orifice,  mul- 
tiplied by  the  distance  gg' .  But  a 
more  simple  way  of  considering 
the  problem,  is  to  assume  that  the  centre  of  gravity  of  the 
part  below  the  slice  A  CB  remains  at  g',  and  hence,  that 
the  change  in  the  position  of  the  centre  of  gravity  has 
been  produced  by  the  transference  of  the  slice  A  CB  to  the 
level  of  the  orifice  F.  Assuming  that  there  are  no  fric- 
tional  resistances,  nor  resistance  from  the  air,  then  the 
whole  energy  will  be  expended  in  producing  the  motion 
of  the  liquid. 
Let 

8  =  the  horizontal  section  of  the  vessel  at  AB^ 

k  =  the  section  of  the  orifice  at  F, 

x  =  the  thickness  of  the  thin  slice  AB, 

h  =  DC  =  the  vertical  height  of  AB  above  the 

orifice, 
i  =  the  time  of  the  discharge  of  a  quantity  of  the 

liquid  equal  to  that  in  the  slice  AJB, 
w  =  the  weight  of  a  unit  of  rolume — say  one  cubic 
inch  of  the  liquid,  and 


168  SOLUTION    OF    PROBLEMS.  f250.J 

t)  =  the  velocity  of  the  discharge. 
Then, 

Sx  —  the  volume  of  the  slice, 
wSx  =  the  weight  of  the  slice,  and 
(wSx)h  —  the  work  accumulated  in  the  slice  when  it  has 
fallen  through  the  height  h. 

The  quantity  which  will  flow  through  the  orifice  in  a 

time  t  will  be 

kvt, 

the  weight  of  which  will  be 

wkvt, 
the  mass  of  which  is 

wkvt 

~' 
and  of  which  the  kinetic  energy,  due  to  the  flow,   will  be 

•     ,  (wkvt}   „ 

1  ±  -  L  $)«  • 

*     g 

hence, 


But  the  weight  wSx  =  wkvt)  and  by  cancelling  and  re 
ducing  we  have 


which  is  the  same  as  that  of  a  particle  falling  freely 
through  a  height  h. 

This  result  is  modified  in  practice  on  account  of  the  re- 
sistance of  the  air,  friction,  and  viscosity  of  the  liquid.  If 
there  is  a  pressure  upon  the  top  of  the  vessel,  we  ascertain 
what  height  of  the  liquid  will  produce  the  same  pressure 
per  square  inch  of  the  upper  surface,  and  add  the  height 
to  the  value  of  A,  given  in  the  problem.  The  height 
which  induces  the  flow  is  called  a  head. 


[250.J  KIXE.TIC   ENERGY.  169 

EXAMPLES. 

1.  If,  in  Fig.  107,  P  =  25  Ibs.,  W=  30  Ibs.,  the  angle 

B  =  60°,  and  J.  —  30°,  determine  which  will  move 
down  the  plane,  P  or  W,  and  how  far  they  will 
move  in  5  seconds. 

2.  Determine  the  relations  between  the  weights  P  and  W, 

and  the  angles  A  and  B  for  equilibrium  in  Fig.  107. 

3.  If  the  angle  A  =  30°  and  B  —  45°,  find  the  relation 

between  P  and  W  so  that  the  acceleration  of  the 
bodies  will  be  £  that  of  a  body  falling  freely. 

4.  In  Fig.  102,  if  the  radius  GA  is  3  feet,  AE  I  foot,  and 

FB  6  inches,  and  the  internal  cylinders  of  the 
same  material,  what  will  be  the  angle  ECg  for 
equilibrium  \ 

{>.  Solve  the  problem  in  Article  248,  when  there  is  a  con- 
stant frictional  resistance  on  the  plane  equal  to 
p  Wcos  A. 

6.  In  the  problem  of  Article  241,  find  the  angle  ECg^ 

where  the  cylinders  E  and  F  have  the  same  diame- 
ter, but  F!  =  2  F2. 

7.  A  vessel  is  filled  with  a  liquid  and  kept  constantly  full ; 

required  the  time  necessary  for  the  discharge  of  a 
quantity  q  from  an  orifice  whose  section  is  k,  the 
distance  of  the  orifice  below  the  surface  of  the  liquid 
being  h  feet. 


CHAPTER  XI. 

CONSTRAINED   EQUILIBRIUM. 

251.  A  body  is  said  to  be  constrained  when  it  is  pre- 
vented from  moving  in  a  particular  direction  on  account 
of  a  point  or  line  of  the  body  being  fixed,  or  on  account 
of  the  interposition  of  a  body  which  is  considered  as  im- 
movable.    By  the  term  fixed  is  not  meant  that  the  resist- 
ance offered  by  a  point,  line,  or  surface,  cannot  be  over- 
come by  any  force,  but  that  it  will  not  be  equaled  by  any 
force  which  may  be  involved  in  the  problem.     The  term 
immovable  is  also  considered  in  the  same  restricted  sense. 
When  only  a  point  of  the  body  is  fixed,  the  body  will  be 
free  to  rotate  about  it  in  all  directions ;  and  when  a  line 
is  fixed  the  body  may  rotate  about  it  or  slide  along  it. 

252.  Normal  Resultant. — If  a  body  be  in  equilibrium 
oil  a  smooth  surface,  the  resultant  of  all  the  forces  which 

act  upon  it  must  be  in  the  direction 
of  the  normal  to  the  surface  and  act 
towards  the  surface.  For,  if  it  be  not 
normal,  it  may  be  resolved  into  two 
components,  one  of  which  will  bo 
normal  and  the  other  tangential,  the 
Plo  110  latter  of  which  would  produce  mo- 

tion. 

253.  Equilibrium  of  a  Body  on  a  Smooth  Inclined 
Plane. — Let  AC  be  the  inclined  plane,  o  the  centre  of  the 
body,  J^the  resultant  of  all  the  forces  which  act  upon  the 
body,  and  fFthe  weight  of  the  body.     Resolve  the  weight 


[254.] 


CONSTRAINED    EQUILIBRIUM. 


171 


W  into  two  components,  one,  cb,  parallel  to  the  plane,  and 
the  other,  ob,  normal  to  it ;  and,  similarly,  resolve  tho 
force  F  into  the  normal  com- 
ponent de,  and  the  component 
oe  parallel  to  the  plane.     The 
excess   of    ob    over  de    need 
not  be   considered,  since   the 
plane  must  resist  it ;  but,  that 
there  shall  not  be  movement 
along  the  plane,  the  compo- 
nent cb  must  equal  oe. 
From  the  figure  we  have 


Fio.  111. 


and 


cb  =  oc  sin  cob 
=  TFsin^l ; 


oe  =  od  cos  doe 
—  Fcoa  fa 

where  <f>  =  doe,  the  angle  between  the  action-line  of  the 
force  and  the  plane,  hence 

j^cos  <f>  —  TTsin  A  ; 

W   . 
.*{  cos  (j>  =  —r,  sm  A. 

254.    Equilibrium  on  a  Rough  Inclined  Plane.  — 

Let  the  notation  be  as  in  the  preceding  article.  The 
amount  of  friction  will  be,  according  to  Articles  107  and 
108,  the  normal  pressure  multiplied  by  the  coefficient  of 
friction,  or, 

fi(ob  —  de) 
=  ji  TFcos  A  — 


If  the  body  is  in  a  state  bordering  on  motion  down  the 
plane,  the  upward  pull  of  F  along  the  plane,  and  the 
friction,  will  equal  the  downward  pull  of  W;  or, 


172  STATICS.  [255.] 

FCOB  (f>+p(  Wvos  A—  Fsm  <£)  =  TFsm  A  ; 

from  which  we  find 

„     sinJ.—  /LICOS  A  TJT 

•F  —  ~    ~L  —    —  :  —  T~  "• 

cos  9  —  fjb  sm  </> 

If  the  value  of  F  is  such  that  the  body  is  in  a  state 
bordering  on  motion  up  the  plane,  the  component  of  F, 
parallel  to  the  plane,  will  equal  the  component  of  W,  also 
parallel  to  the  plane,  plus  the  friction  ;  or 

FCOQ  (f>  =  TFsin  A  +/*(  IT  cos  A—  F&m  <£) 


.  p—  sn  ^-+  /*  cos  ^- 
cos  <  +  A  sin  (> 


255.  Determine  the  conditions  of  equilibrium  of  a 
homogeneous  disc,  having  a  hole  cut  in  it  near  one  edge, 
the  disc  being  placed  vertically  on  an  inclined  plane  which 
is  sufficiently  rough  to  prevent  sliding, 

This  problem  is  essentially  the  same  as  if  one  side  of  the 
disc  were  loaded  with  a  heavier  substance,  or  if,  from  any 

other  cause,  the  centre  of  gravity 
is  not  at  the  centre  of  the  disc. 

Let  g  be  the  centre  of  gravity 
of  the  body.  The  point  of  sup- 
port for  equilibrium  must  be  at 
d  vertically  under  g,  which  will 
also  be  a  point  of  tangency  of 
110  the  circle  and  plane. 

With  c  as  a  centre,  and  radius 

eg,  describe  a  circle  ;  then  if  the  centre  cf  gravity  be  at  g' 
vertically  over  d,  the  body  will  also  be  in  equilibrium.  In 
the  former  position  the  equilibrium  will  be  stable  ;  in  the 
latter,  unstable.  In  the  latter  case,  if  the  body  be  dis 


[256,257.]         CONSTRAINED    EQUILIBRIUM. 


173 


turbed,  it  may  appear  to  roll  up  the  plane,  hut  the  centre 
of  gravity  will  really  be  falling  until  it  assumes  a  position 
of  stable  equilibrium.  If  the  vertical  through  d  falls  out- 
side the  circle  gg',  it  will  not  be  in  equilibrium  in  any 
position  ;  if  tangent  to  it,  the  body  will  be  in  equilibrium 
in  only  one  position. 

256.  To  find  the  inclination  of  the  plane  so  that  th* 
unbalanced  disc  of  the  preceding  problem  will  just  be  on 
the  point  of  rolling  down  the  plane. 

According  to  the  conditions  given  in  the  preceding  arti- 
cle, the  vertical  through  d  must  be  tangent  to  the  circle 
gg' .  Through  c  draw  a  horizontal  line,  and  a  right-angled 


FIG.  113. 


triangle  cgd,  will  be  formed,  in  which  the  angle  cdg,  Fig. 
113,  will  equal  CAB  ;  hence, 

.     ,      eg 

em  A  =  sin  d  =  -^ 
cd 

_  dist.  of  centre  of  gravity  from  centre  of  circle 
radius  of  the  circle. 

257.  A  small  prismatic  bar  rests  upon  two  smooth  in- 
clined planes  ;  determine  the  position  for  equilibrium. 
Let  AJB  be  the  beam,  and  E  its  centre  of  gravity.    The 


174 


STATICS. 


[258,250.] 


reactions  at  A  and  JS  must  be  perpendicular  to  the  respect- 
ive planes :  and  since  these  reactions,  and  the  weight  W, 
are  the  only  forces  which  act  upon  the  body,  their  lines  of 
action  must  all  meet  in  a  common  point.  Hence,  the 
perpendiculars  AD  and  BD  must  meet  the  vertical 
through  E  at  a  common  point.  If  a  string  ADB  be 
attached  to  the  extremities,  A  and  B,  of  the  bar,  and 
hung  on  a  smooth  pin  at  1),  vertically  over  the  centre  of 
gravity  E,  there  will  be  no  tendency  to  slide  on  the  pin, 
and  the  bar  will  remain  at  rest  in  the  inclined  position. 
If  the  inclinations  of  the  planes  are  given,  a  formula  may 
be  found  for  the  inclination  of  the  bar. 

258.  A  prismatic  bar  rests  upon  the  edge  of  a  smooth 
hemispherical  bowl,  one  end  being  against  the  inner  sur- 
face of  the  bowl-  required  the  position  for  equilibrium. 

Let  E  be  the  centre  of  gravity  of  the  bar,  and  C  the 
centre  of  the  bowl.     At  A  the  reaction  will  be  normal  to 

the  surface  of  the  bowl ;  and 
hence,  its  direction  will  coin- 
cide with  the  radius  and  pass 
through  the  centre  C.  At  F 
the  reaction  will  be  perpen- 
dicular to  the  bar ;  hence, 
when  the  radius  through  A 
prolonged,  meets  the  perpen- 
dicular FD,  at  a  point  D  in  a 
vertical  through  E,  the  bar  will  be  in  equilibrium. 

259.  A  weight  W  is  on  the  arc  of  a  vertical  circle  whose 
centre  is  O,  and  is  held  in  position  by  a  weight  P  sus- 
pended from  a  cord,  which  passes  over  a  pin  at  C,  verti- 
cally  over   O  /    required  the  position  for   equilibrium, 
there  being  nofrictional  resistances. 


FIG.  115. 


[360.J 


CONSTRAINED    EQUILIBRIUM. 


175 


The  tension  upon  the  string  will  be  equal  to  the  weight 
P  ;  hence,  the  body  W  will  be  held  by  a  force  equal  to  P 
acting  in  the  direction  cO,  the  force  of 
gravity  acting  vertically  downward, 
and  the  normal  reaction  of  the  curve. 
The  normal  to  the  circle  coincides  with 
the  radius  passing  through  the  point, 
and  hence,  passes  through  the  centre  of 
the  circle. 

Draw  a  vertical  ca  to  represent  the 
weight  TF,  and  ab,  parallel  to  cC,  to  re- 
present P  ;  then  the  position  must  be 
such  that  the  extremity  b,  of  the  line 
ab,  will  fall  on  the  normal  cO.  The 
triangles  abc  and  OOo  are  similar,  hence,  we  have 

W:P::ca:ab 

::OO:  Cc- 


260.  A  particle  W,  attached  to  one  end  of  a  string,  is 
placed  on  the  convex  side  of  a  smooth  parabola,  and  held 
in  position  by  a  weight  P  attached  to  the  other  end  of 
the  string,  the  string  passing  over  a  smooth  pin  at  the 
focus  of  the  parabola  /  required  the  position  for  equili- 
brium. 

Let  the  axis  of  the  parabola  be  vertical,  C  the  focus, 
and  c  the  position  of  the  weight  W.  Construct  the  tri- 
angle of  forces  cba  as  before,  ce  being  the  normal  to  the 
parabola.  Since  ca  is  parallel  to  the  axis  Ce,  the  normal 
will  bisect  the  angle  Cca  ;  hence, 


176 


STATICS. 


[261.] 


Ccb  =  boa ; 
but,  from  the  construction, 

abc  =  Cob ; 
/.  abc  =  boa ; 
therefore,  the  triangle  abc  must  be  isosceles,  and  we  have 

ab  =  ac, 
or, 

W=P, 

which  establishes  a  relation  between  the  given  quantities. 
It  does  not  determine  any  particular  point,  but  it  shows 


a' 


Pro.  117. 


FIG.  118. 


that,  in  order  that  the  weights  shall  be  in  equilibrium  at 
any  point,  they  must  equal  each  other,  in  which  case  they 
will  be  in  equilibrium  at  all  points  on  the  curve. 

261.  Instead  of  a  parabola,  let  the  curve  be  an  ellipse,  tht 
major  axis  being  vertical  and  the  pin  at  the  upper  focus. 
Construct  the  triangle  of  forces  cab  as  before,  ce  being 
the  normal.     Then, 

W'.P'.'.ca-.ab 
\\de\dc. 


[262.]  CONSTRAINED      EQUILIBRIUM.  177 

Let  e  be  the  eccentricity  of  the  ellipse,  then  it  is  shown 
in  Conic  Sections,  that 

de  :  dc  :  :  e  :  1  ; 
hence. 

W:P;:e:l; 
or, 

W 


that  is,  when  the  ratio  of  W  to  P  equals  the  eccentricity 
of  the  ellipse,  they  will  be  in  equilibrium  for  all  positions 
of  W  on  the  curve  ;  and  if  they  have  any  other  ratio  they 
will  not  be  in  equilibrium  at  any  point,  except  at  the  ex- 
tremities of  the  axis,  A  and  JB. 

[In  "  Conic  Sections  "  the  proportion  involving  e,  given  in  this 
article,  may  not  be  given  in  this  form.     If  a  be  the  semi-major 
axis,  and  x  the  abscissa  of  the  point  c  in  reference  to  the  centre, 
then  the  following  equations  are  usually  given,  viz.  : 
the  radius  vector,  de,  =  a  ±  ex; 
the  distance  from  the  focus  to  the  foot  of  the  normal, 
<fo,  =  e(a  ±  ex)  ; 

the  plus  sign  being  used  for  the  distance  dd,  and  tho  minus  sign 
for  ed.    These  values  readily  give 

de  _  e(a  ±  ex)  _ 
dc~    a  ±  ex 

hence,  the  proportion  given  in  the  text.] 

262.  Let  the  curve  be  an  hyperbola,  the  pin  being  in  tht 
upper  focus. 
Proceeding  as  before,  we  find 

W  _ 

"77  —  e'i 

but  in  this  case  e  exceeds  unity  ;  hence  Fr  must  exceed  Pt 
while  in  the  former  case  TF"must  be  less  than  P. 


178 


STATICS. 


[263.] 


263.  Suppose,  in  the  preceding  case,  that  the  pin  over 
which  the  string  passes  is  at  the  centre  of  the  hyperbola. 
Let  a  be  the  semi-major  axis,  b  the  semi-minor  axis,  the 


axis  of  x  horizontal,  and  y  vertical.     The  axis  of  the  hyper- 
bola being  vertical,  the  equation  will  be 

«A?_jy=-«*P.      ...    (1) 

Construct  the  triangle  of  forces  cab,  in  which  ab  will  be 
parallel  to  cC  ';  draw  cD  horizontal,  and  we  have 

W:  P:\ca\db 

::Ce:  Cc; 

which,  according  to  the  principles  of  Conies,  becomes 


or,  observing  that  CD  =  y,  and  DC  —  x,  we  have 

W:  P'.'.fy. 
hence 


which,  reduced  to  an  equation,  gives 


.        .        .    (2) 


[263.]  CONSTRAINED    EQUILIBRIUM.  179 

According  to  Conies  we  also  have 

<?-*!£  (B) 

tr  —          2     •  •  •  •      vv 

a8 

Eliminating  a  and  x  from  these  three  equations,  we  find 
bW 


y  ~ 


. 

W* 


EXAMPLES. 

1.  In  Fig.  Ill,  if  W=  50  Ibs.,  A  =  30°,  /&/=  60°,  what 

must  Fbc  for  equilibrium  ? 

2.  In   Fig.   121,  if   CAB  =45° 

W=  60  Ibs.,  what  must  F 
be,  acting  horizontally,  for 
equilibrium  ?  (The  value  of 
F  may  be  found  from  the 

equation  of  Article  253,  observing  that  <£  will  be 
-  45°.) 

3.  Find  the  value  of  F,  Fig.  Ill,  for  equilibrium  when  it 

acts  parallel  to  the  plane.  (Deduce  it  from  the 
equation  of  Article  253.) 

4.  Similarly,  find  the  value  of  F,  for  equilibrium,  when 

it  acts  horizontally. 

5.  In  Fig.  Ill,  if  A  =  30°,  W=  50  Ibs.,  coefficient  of 

friction  p  =  0.2,  the  angle  $  =  —  5°  ;  what  will 
be  the  value  of  j^so  as  to  just  prevent  motion  down 
the  plane  ;  also  so  that  it  shall  be  just  on  the  point  of 
moving  it  up  the  plane  ? 


180  STATICS. 

6.  In  the  same  figure,  if  A  =  45°,  W  =  100  Ibs.,  /*  = 

0.15,  what  will  be  the  value  of  F,  acting  parallel 
to  the  plane,  so  as  just  to  prevent  motion  down  the 
plane. 

7.  In  Fig.  113,  if  the  diameter  of  the  small  circle  equals 

the  radius  of  the  larger  one,  and  the  remainder  of 
the  larger  one  be  homogeneous,  what  will  be  the 
greatest  inclination  of  the  plane  A  C  that  the  body 
may  remain  at  rest,  there  being  no  sliding  ? 

8.  The  weight  W  being  20  Ibs.,  P  15  Ibs.,  in  Fig.  116, 

the  radius  of  the  circle  2  feet,  the  distance  O  C\  from 
the  centre  O  to  the  pin,  being  4  feet ;  required  the 
angle  OCc  for  equilibrium? 

9.  In  Fig.  116,  if  OC =  Co  —  the  radius  of  the  circle, 

what  will  be  the  relation  of  P  and  W,  and  the  angle 
COc,  for  equilibrium. 

10.  In  Fig.  120,  if  W  =  100  Ibs.,  P  =  25  Ibs.,  b  =  3  feet, 

and  e  =  2,  what  will  be  the  distance  CD  for  equili- 
brium. 


EXERCISES. 

1.  If  a  force,  equal  to  the  reaction  of  the  plane,  be  substituted  for  the 

plane,  will  the  body  remain  in  equilibrium  ? 

2.  If  a  plane  is  perfectly  smooth,  can  a  body  remain  at  rest  on  it  if  the 

plane  has  any  inclination  ?  If  the  body  were  pressed  normally 
against  the  plane,  would  it  remain  at  rest  ? 

3.  In  Fig.  112,  what  will  be  the  position  of  the  disc  so  that  the  poten- 

tial energy  shall  be  greatest,  and  what,  so  that  it  shall  be  least  ?  In 
Fig.  113,  is  the  potential  energy  a  maximum,  a  minimum,  or  in- 
different ? 

4.  Is  the  potential  energy  of  the  bar  in  Fig.  114  a  maximum  or  a  mini 

mum  ?     Similarly,  in  regard  to  the  bar  in  Fig.  115  ? 

5.  When  the  relations  of  Pto  Ware  such  as  to  secure  equilibrium  in 

Figs.  117,  118,  and  119,  will  the  potential  energy  be  a  maximum, 
a  minimum,  or  indifferent  ? 


[263.]  CONSTRAINED    EQUILIBRIUM.  181 

6.  In  Article  263,  when  the  bodies  are  in  equilibrium,  will  the  potential 

energy  be  a  maximum  or  a  minimum  ? 

7.  In  Article  262,  can  the  bodies  be  in  equilibrium  on  the  curve,  if  P 

exceeds  W? 

S.  In  Fig.  118,  if  W  exceeds  P,  can  there  be  equilibrium  at  any  point 
of  the  curve  ? 

9.  In  Fig.  115,  what  is  the  longest  bar  that  can  rest  on  the  bowl  and 

have  one  end,  A,  in  the  bowl. 

10.  If  the  planes  are  equally  inclined  in  Fig.  114,  what  will  be  the  posi 

tion  of  the  bar  when  in  equilibrium  ? 

11.  If  the  plane  AC,  Fig.  114,  were  horizontal  and  perfectly  smooth 

what  must  be  the  position  of  the  bar  for  equilibrium  ? 


CHAPTER  XII. 


ANALYTICAL   METHODS. 

264.  Analytical  Mechanics  is  generally  understood  to 
refer  to  that  system  of  analysis  in  which  the  equations  of 
equilibrium,  or  of  motion, "are  established  in  reference  to 
a  system  of  coordinates,  and  the  results  obtained  by  opera- 
tions upon  those  equations.     The  coordinates  most  com- 
monly used  are  Rectangular  or  Polar. 

265.  Before  establishing  the  general  equations,  we  ob- 
serve that  when  the  forces  are  in  a  plane  they  may  be 


Flo.  122. 


Fio.  123. 


reduced  to  a  single  resultant,  or  to  a  single  couple,  Article 
194.  If  they  are  not  in  one  plane,  let  P,  Fig.  123,  be  a 
force  applied  at  B,  and  ^another  force  applied  at  A.  At 
any  point  A,  of  the  force  f]  introduce  two  equal  and  oppo- 
site forces,  each  equal  and  parallel  to  P,  and  as  they  mutu- 
ally destroy  each  other  they  will  not  change  the  mechanical 
condition  of  the  problem.  Now,  combining  P  at  £  with 
—  P  at  A,  we  have  a  couple,  whose  moment  is 

P.AB-, 
and  the  other  forces,  Fand  +  P,  at  A,  will  have  a  single 


[366,  26?  ] 


ANALYTICAL    METHODS. 


183 


resultant.  A  similar  operation  may  be  pei formed  when 
there  are  several  forces ;  hence,  we  infer  that,  when  the 
forces  are  not  in  a  plane,  they  may  be  reduced  to  a  single 
force  and  a  single  couple.  This  result,  however,  may  be 
proved  from  the  following  equations. 

266.  General  Case. — Forces  may  be  applied  at  any  or 
all  the  points  of  a  body,  and  act  in  all  conceivable  direc- 
tions.    The  angles   which    the 

action -lines  of  the  forces  make 
with  the  axes  may  be  determined 
as  in  Article  157.  The  origin 
of  coordinates  may  be  taken  at 
any  point,  and  the  rectangular 
axes  have  any  position  in  refer- 
ence to  the  body  or  the  forces. 
The  axis  of  x  will  be  considered  FlG  ]24> 

as  positive  to  the  right,  y  posi- 
tive upward,  and  z  positive  in  front  of  the  plane  yx. 

267.  Forces  Resolved. — Let  the   coordinate  axes  be 
rectangular,  and  the  forces  resolved  parallel  to  them. 

Let       f[,  F2,  fl,  etc.,  be  the  forces, 

a^  02,  03,  etc.,  the  angles  which  the  lines  of  the 

forces  make  respectively  with  the  axis 

OX, 

&>  #2>  A?  etc.,  the  corresponding  angles  with  OY, 
7ij  72?  73?  etc.,  the  corresponding  angles  with  OZ, 
Xj  Y~,  Z,  the  algebraic  sum  of  the  components  of 

the  forces  parallel  respectively  to  the 

axes  of  x,  y,  and  0, 
J?,  the  resultant  of  all  the  forces,  and 
a,  5,  and  c,  the  angles  which  it  makes  with  the  axes 

«,  y,  0,  respectively. 


184  STATICS.  [267.] 

The  components  of  the  forces  parallel  to  the  axis  of  x 
will  be,  according  to  Article  156, 

F!  cos  «!,    F2  cos  03,    Fa  cos  03,  etc.,  and 
R  cos  0. 

These  components  constitute  a  system  of  parallel  forces, 
which  are  not  confined  to  one  plane,  bnt  are  simply  paral- 
lel to  the  axis  of  ac.  The  resultant  of  any  two  of  the  forces, 
since  they  are  in  one  plane,  will  be  their  algebraic  sum, 
Articles  190  and  198  ;  or, 

FL  cos  Ox  -f  Fz  cos  03. 

This  resultant  and  a  third  force  will  be  in  one  plane, 
and,  hence,  the  resultant  of  this  resultant  and  a  third  force 
be  their  algebraic  sum  ;  or, 


L  cos  dt  +    z  cos  OH  +    S  cos  Og. 

Continuing  this  process,  we  finally  have  for  the  result- 
ant of  all  the  components  parallel  to  the  axis  of  x, 


X.  '  =  Fiuoaoi+Fz  cos  a^  +  Fs  cos  Og  +  etc.  = 

which  value  must  equal  the  aj-componeut,  of  the  result- 
ant of  all  the  forces  ;  hence 

X  =  R  cos  a  =2Fcos  a. 

Proceeding  in  a  similar  way  with  the  y  and  s-  compo- 
nents, we  have 

T  =  R  cos  b  —  SFcos  13, 

Z  =  R  cos  c  =•  SFaos  y. 
Squaring  and  adding,  we  have 


)2  +  (^^cos  /3)2  +  (2-Fcos  7)2. 


[268.J  ANALYTICAL    METHODS.  185 

The  direction  of  the  axis  may  be  determined  from  the 
equations 

X  T  Z 

cos  a  =  p  ,    cos  o  =  -j5  5    cos  G  =  TP  • 

If  there  is  equilibrium  we  have 

^  =  0; 
/.  X=0;      T  =  0;     Z=0. 

These  equations  are  the  same  as  those  for  concurrent 
forces,  Article  161  ;  hence,  the  tendency  of  a  system  of 
forces,  having  given  intensities  and  directions  of  action, 
to  produce  translation  will  be  the  same  whether  they  con- 
cur or  not,  and  will  be  independent  of  the  points  of  appli- 
cation of  the  forces. 

268.  Moments  of  the  Resolved  Forces. 

Let 

#15  ^i)  s\  be  the  coordinates  of  the  point  of  applica- 
tion of  f[,  and  a  corresponding  notation  for 
the  forces  F^  F^  etc. 

The  component  of  F^  parallel  to  the  axis  of  a?,  will  tend 
to  turn  the  point  of  application  either  to  the  right  or  left 
about  the  axis  of  z  (unless  it  be  in  the  plane  xz),  and  the 
arm  of  the  component  will  be  yl ;  hence  the  moment  will 
be,  Article  164, 

Fv  cos  at .  ?/! ; 

and  the  component  of  F[,  parallel  to  y,  will  also  tend  to 
turn  the  point  of  application  about  the  same  axis,  z,  and 
the  moment  will  be 

PI  cos  & .  xi. 

In  Fig.  125,  the  component  Fv  cos  a,  tends  to  turn  the 


186  STATICS.  [268.] 

point   of    application  right-handed,   and   F^  cos  /8:,  left- 
handed  ;  hence,  the  resultant  moment  in  reference  to  the 
axis  of  z  will  be 


cos  O]  —  a?t  cos  Pi). 
The  quantity  in  the  parentheses  is 
the  arm  of  the  force  F^  as  shown 
in  Article  176.     This  expression  is 
_  equivalent  to 

FIG.  125. 

F\  cos  a\ .  y\  —  FI  cos  ^ .  Xi  j 
and  the  other  forces  give 

7^  cos  03 . 2/2  —  ^2  cos  /92  •  a? 
etc.,  etc. ; 

and  taking  the  sum  of  these  expressions,  we  have 
2Fcoa  a.y  -  SFcos  j3.x ; 

or, 

Xy-Yx-, 

and,  similarly,  in  reference  to  the  axis  of  y,  we  have 

Zx  —  Xz ; 
and,  in  reference  to  the  axis  of  x,  we  have 

Yz  —  Zy. 

Let 

G  =  the  moment  of  the  resultant  couple,  which,  as 

shown  in  Article  171,  may  be  represented 
by  its  axis ; 
d  =  the  angle  which   the  axis   of  the   resultant 

couple  makes  with  the  axis  of  a?, 
e  —  angle  which  it  makes  with  the  axis  of  y, 
f=  angle  which  it  makes  with  the  axis  of  z ; 
then  will  the  component  of  the  resultant  couple,  in  refer- 
ence to  the  axis  of  x,  be 


[269.]  PROBLEMS.  187 

and  similarly  for  the  others.     Hence,  we  have  for  equili- 
brium, 

G  cos  d  =  Zy  —  Yz, 

G  cos  e  =  Xz  —  Zx, 

Gcoaf  =  Yx—  Xy. 

If  the  given  forces  are  in  equilibrium  in  reference  to 
rotation,  we  have 

£  =  0; 
.:Zy=Yz, 
Xz  —  Zx, 
Yx  =  Xy ; 

the  last  of  which  may  be  deduced  from  the  two  preceding 
by  eliminating  z. 

In  many  cases  it  is  advisable  to  find  the  moments  of  the 
forces  directly  instead  of  the  moments  of  the  components. 


Problems. 

269.  A  cord  is  secured  at  the  points  A  and  C,  and  sus- 
tains a  weight  attached  to  it  at  B  /  required  the  tension 
of  each  part  of  the  cord, 
the  weight  of  the  cord  be-     t~*~*h^^---2--------^^*^Tti 

ing  neglected,  and    the  "~~J3T~ 

parts  AB  and  BC  being  (^)w 

unequal.  FIG.  126. 

Let  t  =  the  tension  of  AB, 

t\  =  the  tension  of  BC,  and 
W  =  the  weight  of  the  bod}'. 

Taking  the  origin  of  coordinates  at  any  point  as  By 
x  horizontal,  and  y  vertical,  and  resolving  the  forces 
horizontally  and  vertically,  we  have 


188  STATICS.  [270.J 

X  =  t  cos  BAD  —  tj.  cos  BCD  =  0, 
Y=  t&in  BAD  +  t^\nBCD  —  W=Q. 


There  being  only  two  unknown  quantities,  the  equation  of 
moments  will  be  unnecessary.  If  the  origin  of  momenta 
be  at  B,  all  the  moments  will  vanish. 

From  the  first  of  these  equations  we  have 

t  cos  BAD  —  ti  cos  BCD  ; 

hence,  the  horizontal  components  of  the  tensions  are  equal 
to  each  other.  From  the  last  equation  we  have 

cvsBCD 
- 


which,  combined  with  the  second  equation,  gives 

._  sec  £  AD  w. 

~  tan  BAD  +  tan  B  CD 

_  &QcBCD 

*  ~  tan  BAD  +  tan  B  CD 

270.  A  weight  W  is  attached  to  a  cord  DB  of  given 
length  and  pushed  from  a  vertical  wall  by  a  strut  BA  of 
known  length  •  required  the  tension,  t, 
of  the  string  ',  the  compression,  c,  of  the 
strut,  and  the  upward  push,  F,  on 
the  vertical  wall,  when  there  is  equili- 
brium. 

Let  AD  be  known,  and  the  angles  of 
the  triangle  ABD  computed. 

Take  the  origin  of  coordinates  at  A, 
x  horizontal,  and  y  vertical,  and  posi- 
tive upward.  Resolving  the  forces  acting  at  B,  we  have 
(see  Article  157), 


|271.]  PROBLEMS.  189 


X—  t  cos  (90°  +  <£)  +  c  cos  (270°+  0)4-  TFcos  270°=  0; 
T  =  t  sin  (90°  +  $)  +  c  sin  (270°  +  0)  +  TFsin  270°  =  0  ; 
and  taking  the  moments  of  the  forces  directly,  we  have 

SFa  =  +t.AC+  c.Q-W.BE=Q. 
These  reduced,  give 

—  t  sin  (f>  +  c  sin  0  =  0  ; 

+  t  cos  <f>  —  c  cos  6  —  W=  0  ; 

t.AD&iu  <f>—W.  A£sin  0  =  0. 

AC 

These  equations  solved,  observing  that  sin  <f>  =  -rjv  and 

•  EB     ' 

give 


JSB 

-  ~  AC      > 


c  — 


» 

AD 


In  a  similar  way,  resolving  the  forces  c,  Ft  and  X  at 
and  we  find 


and  EB 


271.  Two  braces,  or  rafters,  secured  at  their  lower  ends 
by  a  horizontal  tie  rod,  carry  a  weight  W  at  the  point 
where  they  meet  /  required  the  vertical  action  at  the  sup- 
ports, the  tension  of  the  horizontal  tie,  and  the  compres- 
sion of  the  braces. 

First,  consider  the  parts  as  rigid,  and  determine  the  re- 
lations between  the  external  forces. 

Let  the  braces  be  equally  inclined,  then  will  D,  verti- 


190 


STATICS. 


1271.) 


cally  under  W,  be  midway  between  A  and  B.  Also,  let 
F!  be  the  vertical  action  of  the  support  at  A,  and  V  that 
at  B.  Take  the  origin  of  coordinates  at  A,  x  horizontal 
and  y  vertical,  and  we  have,  from  Fig.  128, 

X  =  F!  cos  90°  +  Fcos  270°  +  Fcos  90°  =  0  ; 
7  =  F!  sin  90°  +  JFsin  270°  +  V&m  90°  =  0  ; 


and  these  give 


$Fa  = 
which  readily  give 


hence,  each  support  sustains  one-half  the  load  W;  a  result 
readily  deduced  from  the  principle  of  the  lever. 


FIG.  128. 


Secondly,  to  find  the  stresses  which  are  transmitted 
along  the  pieces,  conceive  that  a  plane  EF  is  passed 
through  two  of  the  pieces,  cutting  them  at  the  points  E 
and  F.  Let  the  part  AEF,  between  the  end  A  and  the 
plane  section,  be  removed,  and  let  the  forces  c  and  t, 
equal  to  the  compression  and  tension  which  previously  ex- 
isted, be  substituted.  The  frame  in  the  new  condition 
will  be  in  equilibrium,  the  forces  c  and  t  being  the  unknown 


[272.]  PROBLEMS.  191 

quantities  to  be  determined  from  the  equations  of  equili- 
brium. The  direction  of  action  of  these  forces  is  repre- 
sented by  the  arrow  heads,  Fig.  129. 

The  origin  of  coordinates  remaining  at  A,  we  have 

X  =  Fcos  90°  +  TFcos  270°  +  t  cos  180° 

+  ccos  CAD  =  0, 
T  =  Fsin  90°  +  Wsm  270°  +  t  sin  180° 

+  G  sin  CAD  =  0, 

2Fa  =  V.  AB  -  W.  AD+t  .  0  +  c.  0  =  0  ; 
which  give,  observing  that  V  =  %  W, 
—  t  +  ccos  CAD  —  0, 


and  these  give 

c  =  £  JFcosec  CAD, 

t  =  $WeotCAZ>', 
or, 

G  cos  CAD  =  t, 
c  sin  CAD  =  i  PF; 

from  which  it  appears  that  the  horizontal  component  of 
the  compression  on  A  C  equals  the  tension  of  the  horizon- 
tal tie  AJB,  and 

The  vertical  component  of  the  compression  on  AC 
equals  the  vertical  action  at  A. 

In  a  similar  way  the  stresses  on  any  frame  may  be  de- 
termined, provided  the  plane  section  does  not  intersect 
more  than  three  acting  members.  There  being  three 
equations  for  equilibrium  for  forces  in  one  plane,  two  for 
forces',  and  one  for  moments,  only  three  unknown  quanti- 
ties can  be  determined  from  them. 

272.  Remark.  —  These  solutions  are  given  that  the  stu- 
dent may  learn  the  process  of  establishing  the  general 
equations,  and  not  because  the  solutions  are  the  shortest 


192 


STATICS. 


[273.] 


or  the  most  simple.  The  solutions  here  given  are  appar- 
ent!}7 longer  than  are  necessary,  for  those  quantities  and 
terms  which  reduce  to  zero  need  not  be  entered  in  the 
equations ;  but  it  was  thought  best,  for  practice,  to  enter 
every  force  in  each  equation.  The  Analytical  method  is 
not  used  because  it  makes  the  solution  of  a  particular 
problem  shorter  than  by  special  methods,  but  because  it 
establishes  a  uniform  method  of  procedure,  and  also  fur- 
nishes a  powerful  method  <yf  investigation  in  more  dijji- 
cv2t  problems. 

273.  The  preceding  problem  may  be  solved  by  the 
following  shorter  method.     Let  Cb  represent  the  weight  W, 

and  ba  be  drawn  parallel  to  B  6Y, 
and  ad  horizontal ;  then  will  ab 
represent  the  compression  on 
CB)  a  C  •=•  ab  =  the  compres- 
sion on  CA,  and  ad  the  tension 
on  AB.  Cd  =db  -^W:  hence, 
FIG.  130.  .  we  have 


or, 


and 


or, 


ad  =  Cd  cot  Cad ; 

t  =  |  TFcot  CAB ; 

a  C=  CWcosec  Cad, 

o  =  %  TFcosec  CAB. 


EXAMPLES. 

1.  In  Fig.  126,  if  W=  100  Ibs.,  AC  =  5  ft.,  AB  =  BC 

=  3  f t.,  what  will  be  the  tension  of  the  cord  ? 

2.  In  the  same  figure,  if  the  length  of  the  cord  is  10  ft, 

AC,  5  ft,  what  must  be  the  length  of  AB  that  the 
tension  of  it  shall  be  twice  that  of  BC\ 


[273.]  PROBLEMS.  193 

3.  If  AB  be  horizontal,  in  Fig.  126,  and  BC  inclined,  find 

the  tension  on  AB  and  BC, 

4.  In  Fig.  128,  if  AB  is  20  ft.,  DC  10  ft,  and  TF500  Ibs., 

find  the  tension  on  AB  and  the  compression  on  AC. 
6.  Required  the  inclination  of  the  rafters,  so  that  the  stress 

on  each  will  equal  the  weight  W. 
6.  Required  the  inclination  of  the  rafters  so  that  the  stress 

on  AB  shall  equal  the  weight  IF". 


EXEECISES. 

1.  In  Fig.  122,  if  the  resultant  of  F  and  Fi  equals  R,  and  another  force 

equal  to  jR  be  introduced  at  the  point  of  concurrence,  and  another 
force  equal  and  opposite  to  R  acting  on  another  part  of  the  body 
be  also  introduced,  will  the  four  forces  have  a  single  resultant  ? 

2.  If  the  force  R,  at  the  concurrence  of  F  and  FI  be  removed,  will  the 

three  remaining  forces  have  a  single  resultant  ? 

3.  In  Fig.  125,  if  the  point  of  application  of  the  force  be  in  the  second 

angle,  both  components  will  turn  the  point  of  application  right- 
handed  ;  does  the  expression  for  the  moments  need  to  be  modified 
for  this  case  ? 

4.  In  Fig.  126,  can  the  cord  be  drawn  into  a  straight  line  when  the 

weight  B  is  upon  it  ? 

5.  If  the  parts  of  the  cord  .4 Sand  BC  are  vertical,  what  will  be  the 

tension  on  each  part  ? 

6.  If  the  points  of  support  A  and  C  are  not  in  the  same  horizontal,  while 

the  inclination  of  the  parts  AB  and  BC  remain  the  same,  will 
the  tension  remain  the  same  ? 

7.  If  the  weight  be  moved  to  and  fro  on  the  cord,  in  what  curve  will  be 

the  locus  of  the  point  B  ? 

8.  In  Fig.  128.  if  the  lower  ends  of  the  braces,  A  and  J9,  be  moved  to- 

wards each  other,  will  the  compression  on  the  braces  be  increased 

or  decreased  ?    Will  the  thrust  at  the  lower  ends  be  changed  ? 
0.   What  will  be  the  effect  upon  the  stresses  if  the  apex  be  lowered  by 

lengthening  the  cord  AB? 
10.  If  a  vertical  strut  CD  be  placed  under  C,  will  there  be  any  stress  on 

the  rafters  ? 


CHAPTER  XIII. 


STRENGTH    OF   BAKS   AND   BEAMS. 

274.  Strength  of  Prismatic  Bars. — It  has  been  ob 
served,  Articles  129  and  130,  that,  if  a  solid  be  pulled  in 
the  direction  of  its  length,  it  will  be  elongated.     We  also 
know  that  if  the  pulling  force  be  sufficiently  great,  the 
bar  will  be  broken. 

The  strength  of  solids  is  determined  by  experiment. 
The  most  common  way  of  doing  it  is  to  take  a  bar  whose 


FIG.  132. 


cross-section  is  somewhat  larger  than  the  section  which  it 
is  desired  to  test,  and  then  turn  down  a  portion  near  the 
middle  of  the  piece  to  an  exact  cylinder  of  definite  dia- 
meter. The  piece  is  then  placed  in  a  powerful  machine 
and  pulled  asunder.  The  total  pulling  force  being  mea- 
sured, the  stress  per  unit  of  section  is  easily  computed. 
By  experimenting  upon  pieces  of  different  diameters  and 
lengths,  and  with  different  materials,  certain  general  facts 
or  laws  have  been  determined. 


1275,276.]     STRENGTH  OP  BARS  AND  BEAMS         195 

275.  Results. — The  strength  of  prismatic  bars  varies 
directly  as  their  cross-section.     This  is  the  law  assumed 
in  practice,  and  it  appears  to  be  nearly  evident  without 
experiment ;  but,  like  all  other  physical  laws  pertaining 
to  solids,  it  is  not  rigidly  exact.     It  is  found  that,  when 
the  reduced  portion  ab  is  very  short,the  piece  is  stronger 
than  when  it  is  longer,  but  no  law  for  this  increase  of 
strength  is  known.     Bars  of  small  cross-section  generally 
appear  to  be  stronger  in  proportion  than  those  of  larger 
cross-section,  but  the  difference  may  be  more  apparent  than 
real,  for  it  may  be  due  to  the  difficulty  of  producing  a  pull 
exactly  in  line  with  the  axis  of  the  larger  piece,  and  cross- 
strains  will  generally  break  a  piece  more  easily  than  a 
direct  pull.     If,  however,  the  pieces  are  made  small  by 
being  so  forged  before  they  are  turned,  instead  of  being 
turned  down  from  larger  pieces,  the  difference  in  strength 
may  be  due  to  the  process  of  manufacture. 

When  a  piece  is  elongated,  its  cross-section  is  contracted. 
When  the  material  is  ductile,  the  contraction  may  be  con- 
siderable before  fracture  takes  place.  Thus,  in  some  ex- 
periments upon  wrought  iron,  the  cross-section  has  been 
reduced  in  this  way  to  one-half  of  its  original  area. 
WTien  the  part  ab.  Fig.  132,  is  very  short,  the  contraction 
is  much  less  than  when  it  is  longer.  The  shoulders  of  the 
larger  portion  appear  to  aid  in  resisting  the  contraction 
of  the  smaller  part,  and  this  may  account  for  the  increase 
of  strength  above  referred  to. 

276.  Modulus  of  Tenacity. — The  pull,  in  pounds, 
which  will  break  a  bar  whose  cross-section  is  one  square 
inch,  is  a  measure  of  the  tenacity  of  the  nMterial  and  is 
called   THE  MODULUS   OF   TENACITY.     It  is  represented 


196  STATICS.  [277-279.] 

Mean  Values  of  T. 

For  wrought  iron 45,000  Ibs.  to    60,000  Ibs 

For  steel 70,000    «   to  190,000    " 

American  cast-iron 20,000    "   to    45,000    " 

English  cast-iron 13,000    "   to    25,000    « 

Ash  (English) 15,000    "   to    17,000    " 

Oak  (English) 9,000    "   to    15,000    " 

277.  Formula  for  the  Tenacity  of  Solids. 

Let 

T  =  the  modulus  of  Tenacity, 
P  =  the  pulling  force, 
iS  =  the  area  of  the  cross-section ; 

then  we  have 


278.  Strength  of  Beams. — In  order  to  make  a  formula 
for  the  strength  of  beams,  it  is  necessary  to  know  the  law 
of  action  of  the  forces  within  the  beam.     Within  the  elas- 
tic limits  this  law  is  quite  perfectly  known,  but,  when  a 
state  bordering  on  rupture  is  reached,  the  law  of  action  is 
quite  complex,  and  is  not  accurately  known.    But  the  same 
law  of  resistance  is  assumed  for  both  cases,  and  the  error, 
if  any,  which  results  from  the  assumption  is   corrected 
by  means  of  a  constant  factor. 

279.  Law  of  Resistance. — When  a  beam  is  fixed  at 
one  end  and  loaded  at  the  free  end,  as  in  Fig.  133,  the 
fibres  on  the  upper  side  are  elongated  and  on  the  lower 
side,  compressed,  and  there  is  a  surface  between  the  ex- 
tended and  compressed  elements  which  is  neutral,   and 
is  called  a  neutral  surf  ace.     It  is  assumed  that  the  stress 
on  the  fibres  varies  directly  as  their  distance  from  the 


1380-281.]  STRENGTH    OF    BARS    AND  BEAMS. 


197 


neutral  surface.     The  intersection  of  the  neutral  surface 
with  a  vertical,  Longitudinal  plane  is  called  the  neutral  axis. 


Pio.  133. 


280.  Modulus  of  Rupture. — The  Modulus  of  Rup- 
ture is  the  stress  at  the  instant  of  rupture  on  a  unit  of 
area  of  the  cross-section  which  is  most  remote  from  the 
neutral  axis.     It  is  represented  by  R. 

281.  Expression  for  the  Moment  of  Resistance  of 
Rectangular  Beams. — In  Fig.  133,  let  pr  be  the  neutral 
axis.     Take  In  =  mo  =  ab  =  dc  =  R.      Pass  the  planes 
po  and  pc,  then  will  the  wedge  pq  -  noml  represent  the 
total  pulling  stress,   and  pq-abcd  the   total   pushing  or 
compressive  stress. 

Let 

b  =  Im  =  the  breadth  of  the  beam,  and 
d  =  la  =  its  depth  ;  then 

are&plmq,  and 

—  the  volume  of  the  upper  or  lower  wedge. 
The  moment  of  this  resistance  equals  the  total  resistance 
into  the  distance  of  the  centre  of  gravity  of  the  wedge 
from  the  neutral  surface.     The  centre  of  gravity  of  the 


198  STATICS.  [282.] 

wedge  is  at  the  same  distance  from  the  neutral  surface  as 
that  of  the  triangle  pnl  ;  that  is  %pl  ;  hence,  the  arm  of 
the  resistance  is  f  pi  =  \  x  \d  =  ^d  and  the  moment  is 


and  the  total  moment  of  resistance  of  both  the  upper  and 
lower  stresses  will  be  twice  this  value,  01 


This  expression  is  true  for  all  rectangular  beams  whose 
sides  are  vertical,  whether  the  beam  be  fixed  at  one  end, 
supported  at  its  ends,  or  otherwise,  and  whether  the  beam 
be  loaded  at  one  point,  several  points,  or  uniformly  loaded. 

282.  Problems.  —  1.  Let  a  rectangular  learn  be  fixed  at 
one  end,  to  find  a  load  P.  placed  on  the  free  end,  which 
will  just  break  the  beam. 

In  Fig.  133,  let  I  =  pr  =  the  length  of  the  beam  ;  then 
the  moment  of  P,  in  reference  to  the  fixed  end  as  the 
origin  of  moments  will  be 

Pi; 

and  this  must  equal  the  moment  of  stress  ;  hence, 
PI  = 


From  this  equation  we  find 


frorr  which  the  value  of  R  may  be  easily  computed  when 
the  breaking  weight  P  and  the  dimensions  of  the  beam 
are  known.  The  value  of  R  has  been  determined  for  a 
variety  of  substances,  and  the  results  given  in  tables  in 


[282.]         STRENGTH    OF  BARS  AND    BEAMS.  199 

works  on  the  Resistance  of  Materials.  It  is  a  remarka- 
ble fact  that  the  value  of  R  is  not  the  same  as  T  for  any 
substance. 

Equation  (1)  is  true  for  any  stress  less  than  that  which 
will  break  the  beam.  It  is  customary,  in  practice,  to  use 
a  fractional  part  of  R,  called  the  factor  of  safety,  when 
beams  are  to  be  proportioned.  About  £  to  \  of  R  is  used 
for  wrought  iron,  -fa  for  wood,  and  \  to  \  for  cast  iron, 
which  values  give  for  the  safe  value  of  R  about 

10,000  Ibs.  to  12,000  Ibs.  for  wrought  iron, 
1,000  Ibs.  to    1,200  Ibs.  for  wood, 
5,000  Ibs.  to    6,000  Ibs.  for  cast  iron,  and 

12,000  Ibs.  to  20,000  Ibs.  for  steel. 

Any  one  of  the  quantities  of  equation  (1)  may  be  found 
when  all  the  others  are  given. 

2.  Let  the  beam  be  fixed  at  one  end  and  uniformly 
loaded. 

Let  W  equal  the  total  load  ;  then  will  the  lever-arm  of 
the  load  be  the  distance  from  the  fixed  end  B  to  the  centre 
of  gravity  of  the  load,  which 
is  \l ;  hence,  the  moment  is 


and  the  equation   for  equili- 
brium becomes 

FlG-  134' 


3.  Let  the  beam  be  supported  at  its  ends  and  loaded  in 
the  middle. 

The  point  where  a  beam  is  most  liable  to  break  is  called 
the  dangerous  section.  If  the  beam  at  this  section  is 
strong  enough  to  support  the  load  the  beam  will  not  break. 
In  this  problem  the  dangerous  section  is  at  the  middle  of 


200  STATICS.  [282.] 

the  length;    hence,  the  equation  of  moments  should  be 
established  for  this  point. 

The  reaction  of  each  support  will  be  \P,  and  the  arm 
of  this  force,  in  reference  to  the  middle  point  of  the  beam 
as  an  origin  of  moments,  is  %l,  and  the  moment  is 

and  the  equation  becomes 


4±  Let  the  beam  le  supported  at  its  ends  and  uniformly 
loaded. 

Each  support  will  sustain  one  -half  the  load,  or 


which  acts  upward  and  tends  to  turn  the  beam  one  way. 
The  arm  of  this  force  is  %l  in  reference  to  the  centre  of  the 
beam  taken  as  the  origin  of  moments,  and  its  moment  will  be 

±W.&  =  iWl. 

The  load  between  the  middle  of  the  beam  and  one  of  the 
supports,  which  is  also  ^  TF",  acts  downward,  tending  to 
turn  the  beam  in  the  opposite  direction  to  that  of  the 
former  force  ;  and  the  arm  of  this  force  is  the  distance 
from  the  origin  of  moments  to  the  centre  of  gravity  of 
this  load,  which  is~%l  ;  hence  its  moment  is  —  ^  W,^l  ;  and 
the  resultant  moment  will  be 


and  the  equation  for  equilibrium  becomes 
i  Wl  = 


[282.]        STRENGTH  OF  BARS  AND  BEAMS.  201 

EXAMPLES. 
[It  is  best  to  reduce  all  the  linear  dimensions  to  inches.] 

1.  A  rectangular  beam,  whose  depth  is  8  inches,  length  8 

feet,  It  =  1,400  Ibs.,  is  supported  at  its  ends;  re- 
quired the  breadth  so  that  it  will  carry  500  Ibs.  per 
foot  of  its  length.  Ans.  b  =  3^  inches. 

2.  A  rectangular  beam  is  fixed  at  one  end  and  is  required 

to  carry  1,000  Ibs.  at  the  free  end;  if  £='8  feet, 
R  =  1,200  Ibs.,  and  the  depth  is  four  times  the 
breadth,  required  the  breadth  and  depth. 

3.  A  rectangular  beam,  whose  length  is  12  feet,  breadth 

2  inches,  modulus  of  rupture  10,000  pounds,  is  sup- 
ported at  its  ends  ;  required  the  depth  so  that  it  will 
support  8,000  pounds  placed  at  the  middle  of  the 
beam. 

4.  If  a  beam  whose  length  is  10  feet,  breadth  4  inches, 

and  depth  9  inches,  is  supported  at  its  ends,  and 
broken  by  a  weight  of  20,000  pounds  placed  at  the 
middle  of  its  length,  what  is  the  value  of  R  \ 

5.  A  beam  whose  breadth  is  1^  inches,  depth  3£  inches, 

is  supported  at  its  ends  and  loaded  at  the  middle 
with  10,000  pounds ;  what  must  be  its  length  so 
that  the  stress  on  the  outermost  fibres  shall  be  20,000 
pounds  ? 

6.  A  beam  whose  length  is  15  feet,  breadth  6  inches,  and 

depth  12  inches,  is  supported  at  its  ends ;  required 
the  load,  uniformly  distributed,  which  it  will  safely 
support,  calling  the  safe  value  of  R  12,000  pounds. 

7.  An  iron  beam,  two  inches  square,  projects  horizontally 

from  a  wall ;  what  must  be  its  length  to  break  it- 
self at  the  wall,  the  value,  of  R  being  30,000  pounds; 
and  the  weight  of  the  material  being  i  pound  per 
cubic  inch  I 


CHAPTER  XIV. 

MOTION   OF   A   PARTICLE   ON   AN   INCLINED   PLANE. 

283.  To  faid  the  formulas  for  the  movement  of  a  parti- 
cle down  an  inclined  plane  by  the  action  of  gravity,  all 
resistances  being  neglected. 

Let 

A  C  be  the  plane, 
W  =  the  weight  of  the  particle, 
F=  the  component  of  the  force  of  gravity  par- 
allel to  the  plane, 
8  =  the  distance  over  which  the  body  moves  in 

the  time  t, 

v  =  the  velocity  acquired  in  the  time  £,  and 
<£  =  CAB  =  the  angle  of  elevation  of  the  plane. 

The  force  of  gravity  resolved  parallel  to  the  plane  gives 

F=  TFsin  <£, 

which,  being  constant,  will,  ac- 
cording to  Article  60,  produce  a 
constant  acceleration,  the  value  of 
which  may  be  found  by  Article 
86,  and  is 


F 


=  g  8m 


[284.]  MOTION  OF   A  PARTICLE.  203 

This  value  of  f,  in  the  equations  of  Article  24,  gives 
v  =  gt  sin  <f>,          .         .         .     (1) 
=  V2gs  sin  <£,     .         .         .     (2) 


Let  «  =  AC,  then,  from  the  figure,.  we  have 

*sin  <f>  =  JSC, 
which,  substituted  in  equation  (2),  gives 


which  equals  the  velocity  acquired  in  falling  through  a  ver- 
tical height  £>C.  Since  a  similar  result  will  follow  from  a 
plane  having  any  inclination,  and  hence,  from  several  suc- 
cessive planes  having  different  inclinations,  it  follows  that 
the  velocity  acquired  by  a  particle  in  falling  from  one 
point  to  another  when  there  are  no  resistances,  will  be  in- 
dependent of  the  path  over  which  it  moves,  and  will  equal 
the  velocity  acquired  in  falling  vertically  through  a  height 
equal  to  the  height  of  one  point  above  the  other. 

This  result  also  follows  directly  from  the  principles  of 
Kinetic  Energy,  for  the  energy  imparted  by  gravity  equals 
the  weight  of  the  body  into  the  height  through  which  it  acts. 

The  formulas  of  this  article  may  be  deduced  from  those 
of  Article  247,  by  supposing  that  one  of  the  bodies  van- 
ishes. 

284.  Initial  Velocity.  —  If  the  body  is  projected  down 
the  plane,  or,  in  other  words,  has  an  initial  velocity  v<>  at 
the  instant  t  begins  to  be  reckoned,  we  have, 
v  —  v0  +  gt&h}  0, 
s  =  v0t  +  \g&  sin  (f>  ; 
«*  =  VQ  -f  2gs  sin  0, 


204: 


KINETICS. 


[285,286.] 


and,  if  the  body  be  projected  up  the  plane  with  a  velocity 
v0}  we  have 

v  =  v0  —  gt  sin  <£, 
8  •=.  v0t  —  \g&  sin  </>. 
q?  =  v<?  —  2gs  sin  (ft, 

Problems. 

285.  The  times  of  descent  down  all  chords  of  a  vertical 
circle,  which  pass  through  either  extremity  of  a  vertical 
diameter,  are  the  same. 

Let  ACB  be  the  circle,  and  AC  any  chord  drawn 
through  A.  According  to  equation  (4)  of  Article  283,  the 
time  of  descent  down  A  C  will  be 


t=     /VAC  ^ 
V  a  sin  <£>  * 


g 

Draw  CB,  then  will  the  angle 
ACB  be  right,  for  it  is  inscribed  in 
a  semicircle.  The  angle  </>  is  the  com- 
plement of  CAB ;  hence, 

AC 

sin  <r>  =  cos  A  =    .  n  . 
AB* 

which,  substituted  in  the  preceding  equation,  gives 


which  is  not  only  constant,  but  is   the  time   of  falling 
vertically  through  the  diameter  AB. 

286.  Find  the  straight  line  from  a  given  point  to  a 
given  inclined  plane,  down  which  a  body  will  descend  in 
the  least  time. 


[287.J 


MOTION    OF    A   PARTICLE. 


Let  A  be  the  point  and  BC  the  plane.  Through  A  draw 
a  vertical  line  A  O,  and  on  it  take  O  at  such  a  point  that 
a  circle  described  with  the  radius  OA  shall  be  tangent  to 
BC;  then  will  the  chord  AD,  drawn  from  the  upper  ex- 
tremity A  of  the  diameter,  to  the  point  of  tangency  D,  be 
the  required  line.  For,  all  other  lines  drawn  from  A  tc 
the  right  line  will  be  secants  of  the  circle,  and,  accord- 
ing to  the  preceding  article,  the  times  of  descent  down 


the  internal  portions  will  be  the  same,  and  hence,  the  time 
of  descent  down  the  whole  line  will  be  greater  than  that 
down  AD. 

[To  find  the  centre  of  the  circle,  draw  a  horizontal  line  AB, 
which  line  will  be  tangent  to  the  circle,  and  bisect  the  angle  B  by 
the  line  OB ;  the  point  of  intersection,  0,  of  the  lines  A  0  and 
BO,  will  be  the  centre  required.  From  the  property  of  tangents, 
BA  will  equal  BD.] 

287.  To  find  the  straight  line  of  quickest  descent  from 
one  circle  to  another,  the  centres  of  the  circles  being  in  the 
same  vertical  line,  and  one  tangent  to  the  other  internally. 

Let  A  be  the  point  of  tangency.    Draw  any  secant  ADC} 


206 


KINETICS. 


[230.] 


and  draw  BC.  On  a,B  as  a  diameter,  draw  the  circle 
zbJ2,  and  draw  the  chord  ab.  The  angles  C  and  b  will  be 
right,  for  each  is  inscribed  in  a  semicircle,  and  ab  will  be 
parallel  to  DC.  The  time  of  descent  from  DtoC  will  be 
the  same  as  from  a  to  b,  which  is  the  same  as  from  a  to  B ; 
hence,  the  time  of  descent  from  any  point  of  the  circle 
whose  centre  is  O,  down  the  external  portion  of  the  secant 
drawn  from  A  through  a  point  D  to  the  external  circle,  is 
constant  and  equal  to  that  down  the  vertical  aB.  If  D 
is  a  given  point  on  the  circle  ADa,  then  DO  will  be  the 
line  of  quickest  descent.  If  the  common  tangent  is  at  the 
lowest  point  of  the  circles,  at  B^  and  C  the  given  point ; 
draw  the  secant  JBC,  and  Cb  will  be  the  required  line. 

288.  To  find  the  straight  line  of  quickest  descent  from 
a>  given  circle  to  a  given  line. 

Let  AEB  be  the  circle,  and  CD  the  line.  Through  A 
draw  the  horizontal  line  A  C,  make  CD  equal  to  A  C,  and 
draw  AD ;  the  line  required  will  be  ED. 


FIG.  136. 


289.  To  find  the  line  of  quickest  descent  from  a  straight 
line  to  a  given  circle. 

Let  CD  be  the  line  and  AEB  the  circle.  Draw  the 
horizontal  tangent  A C\  make  CD  —  AC ;  draw  DA,  and 
DE  will  be  the  required  line. 


[290.,  291.]        MOTION    OF    A    PARTICLE.  207 

290.  Find  the  time  of  descent  down  a  rough  inclined 
plane. 

Let  the  forces  be  resolved  as  in  Fig.  131,  and  fi  be  the 
coefficient  of  friction.     The  normal  pressure  will  be 

N  =  Wcosfa 
and  the  resistance  due  to  friction  will  be 

At  TFcos  <j> ; 
hence,  the  effective  moving  force  will  be 

TFsin  <f>  —  p  TFcos  (f> 
•    =  (sin  $  —  p  cos  <£)  TF; 
and  the  constant  acceleration  will  be 

f=.  (sin  <j>  —  /u.  cos  <f>)g  j 

which,  substituted  in  the  equations  of  Article  24,  gives 
v  =  (sin  <f>  —  fj,  cos  <j>)gt,  ....  (1) 
v  =  l/2(sin  <£  —  JJL  cos  <J>)gs,  ...  (2) 
8  =  ^(sin  <j)  —  fj,  cos  fygfi.  ...  (3) 

291.  Approximate  Formulas. — Substitute  the  values 
of  the  sine  and  cosine  in  the  preceding  value  of  f,  and  we 
have 

7         AB\ 

LL \n  ' 


but,  when  the  angle  A  i§  small,  AB  may  be  considered  as 
equal  to  AC,  and  BC  -=-  AC  will  be  the  grade  (a  term 
which  is  in  common  use  in  Railroad  Engineering).  Let  7 
be  the  grade,  and  the  formula  becomes 


and  we  have 

v  =  (7  ~  tifft,      .....     (1) 


v  =     2(y  -  p)gs,      ....     (2) 

(3) 


208  KINETICS. 

292.  Formulas  adapted  to  the  movement  of  cars  on 
planes  of  small  inclination. 

The  grade  is  commonly  given  in  feet  per  mile.  Let  h 
be  the  elevation  per  mile,  then 

A 

7  ~  5280  ' 

and  the  friction  may  be  taken  at  7.5  Ibs.  per  ton  gross  ; 
hence, 

_Y.5_      J_ 
^  ~  2240  ~~  300  ' 
and  the  formulas  become,  with  sufficient  accuracy, 


(2) 


.      .   (3) 

If  the  body  has  a  velocity,  the  distance  it  will  move  on 
a  horizontal  plane,  in  being  brought  to  rest,  may  be  found 
by  making  h  =  o,  in  the  second  of  these  equations,  and  the 
time  of  the  movement,  from  the  first,  after  making  A  =  o 
in  that  equation  ;  hence  we  will  have  for  this  case 

t  =  9fr  (nearly),    ....     (4) 
8  =  4r.6v2  (very  nearly).  .         .     (5) 

EXAMPLES. 

1.  A  smooth  inclined  plane  is  100  feet  long;  what  must 
be  its  inclination  that  the  time  of  descent  of  a  parti 
cledown  it  shall  be  5  seconds? 


[292.]  MOTION    OF    A    PARTICLE.  209 

2.  A  body  is  projected  up  a  smooth  plane  whose  slope  (that 

is  its  inclination  with  the  horizontal)  is  45  degrees, 
with  a  velocity  of  50  feet  per  second  ;  find  its  posi- 
tion at  the  end  of  3  seconds ;  five  seconds ;  ten 
seconds. 

3.  A  body  starts  from  rest  at  the  top  of  an  inclined  plane 

whose  length  is  100  feet,  and  height  20  feet ;  with 
what  velocity  must  a  body  be  projected  up  the  plane 
from  its  foot  in  order  to  meet  the  former  one  at  the 
middle  of  the  plane  ? 

4.  Find  the  line  of  quickest  descent  from  a  point  with- 

out a  circle  to  the  circumference  of  the  circle. 

5.  A  train  of  cars  starts  from  rest  on  an  inclined  plane 

and  runs  down  it  by  the  force  of  gravity  only ;  the 
grade  being  40  feet  to  the  mile  and  the  coefficient  of 
friction  -j-J-g-,  what  will  be  its  velocity  at  the  end  of 
one  mile  ? 

d  A  car  starts  from  the  upper  end  of  an  inclined  plane 
whose  length  is  one-half  of  a  mile,  and  grade  50  feet 
per  mile,  coefficient  of  friction  ^J^,  and  runs  down 
the  plane  ;  how  far  will  it  go  on  a  succeeding  hori- 
zoutal  plane  before  being  brought  to  rest,  the  coeffi- 
cient  of  friction  remaining  constant  ?  <$l*^-~' 

7.  In  the  preceding  example,  required  the  time  of  the 

movement  down  the  plane;  also  the  time  oil  the 
horizontal  plane. 

8.  Required  the  height  of  a  plane  which  is  1,200  feet  long, 

so  that  a  car  starting  at  rest  from  the  upper  end  of  it, 
and  running  down  shall  go  just  1,000  feet  on  a  suc- 
ceeding horizontal  plane,  the  coefficient  of  friction 
being  TJT. 

9.  Suppose  that  there  is  an  available  height  of  25  feet  for 


210  KINETICS.  [292.J 

an  inclined  plane,  required  its  length  so  that  a  body 
descending  it  shall  go  just  800  feet  on  a  succeeding 
horizontal  plane,  the  coefficient  of  friction  being 


[The  examples  from  5  to  9  inclusive  may  be  solved  by  the  for- 
mulas of  the  last  article.  The  resistance  of  the  air,  and  the 
effect  of  the  rolling  mass  in  the  wheels,  is  neglected.  In  the 
formulas  of  the  preceding  article,  t  is  in  seconds,  v  in  feet  per 
second,  a  id  *  in  feet.] 


CHAPTER  XV. 

PROJECTILES. 

293.  A  Projectile  is  a  body  thrown  into  space   and 
acted  upon  by  gravity.     It  is  important  in  the  art  of  Gun- 
nery ;  but  the  solution  of  problems  in  that  Art  pertain- 
ing to  the  flight  of  the  projectile,  to  be  of  any  value,  in- 
volves the  resistance  of  the  air,  and  this  element  so  com- 
plicates the  problem  that  its  solution  will  not  be  attempted 
in  this  work.     Some  interesting  properties  may,  however, 
be  easily  determined  by  assuming  that  the  body  moves  in 
a  vacuum ;  and  it  will  be  found  that  the  formulas  thus 
deduced  are  of  practical  value  in  the  science  of  Hydro- 
dynamics. 

294.  The  path  of  a  projectile  is  the  line  which  it  de- 
scribes.    This  assumes  that  the  body  is  a  particle,  but  if 
the  size  be  considered,  we  would  say  that  it  is  the  line  de- 
scribed by  the  centre  of  its  mass. 

The  horizontal  range  of  a  projectile  is  the  distance 
AB,  Fig.  139,  from  the  point  A,  where  the  body  is  pro- 
jected, to  the  point  J?,  where  the  path  crosses  the  horizon 
tal  plane  passing  through  A. 

295.  The  path  described  by  a  projectile  in  a  vacu- 
um is  a  parabola. 

Let  A  be  the  point  from  which  the  projection  is  made, 
AB  the  line  of  projection,  and  v  the  velocity  of  projection. 
In  a  certain  time,  t,  the  body  will  have  moved  a  certain 
distance,  A£,  under  the  action  of  the  impulse  only; 
hence, 


212 


KINETICS. 


1296.) 


AB  =  vt. 

In  the  same  time  it  would  have  moved  a  distance  AD  = 
BC  under  the  action  of  gravity  only  ;  or 


When  both  motions  are  simultaneous,  it  will  be  at  the 
point  Cj  at  the  end  of  the  time  t.  Eliminating  t  from 
tnese  equations,  gives 

A&  =  —  AD. 
9 

Let  h  =  the  height  through  which  a  body  must  fall  in 


FIG.  137. 


FIG.  138. 


vacuo  to  acquire  a  velocity  v  ;  then,  according  to  Arti- 
cle 72, 


which,  substituted  in  the  preceding  equation,  gives 


which  is  the  equation  of  a  parabola.  The  line  AB  is  a 
tangent  to  the  curve,  and  AD  is  a  diameter. 

296.  To  find  the  velocity  at  any  point  of  the  path. 

Let  A,  Fig.  138,  be  the  point.  The  velocity  will  be 
the  same  as  if  the  body  had  fallen  a  height  h  ;  but  4A, 


[297,298.]  PEOJECTILES.  213 

according  to  the  last  equation,  is  the  parameter  of  the 
parabola  to  the  diameter  AD.  Let  EG  be  the  directrix 
of  the  parabola,  then  will  the  velocity  be  equal  to  that 
due  to  a  fall  through  a  vertical  height  EA  ;  or, 

± 


297.  To  find  the  position  of  the  focus.     Let  F  be  the 

focus,  and  AL  be  drawn  horizontally.  According  to  a 
property  of  the  parabola,  the  tangent  AB  makes  equal 
angles  with  the  diameter  AD  and  the  line  AF  drawn  from 
the  focus  to  the  point  of  tangency  ;  hence, 

BAF=  JAD 

W°  -BAL; 


therefore  the  angle  between  the  tangent  to  the  path,  and 
the  line  from  the  point  of  tangency  to  the  focus,  is  the 
complement  of  the  angle  of  elevation  of  projection. 

The  distance  AF=  EA,  is,  according  to  the  preceding 
article, 

•y2 


hence,  when  the  velocity  of  projection  and  the  angle  of 
elevation  are  given,  the  focus  can  be  found. 

298.  To  find  the  equation 
of  the  path  when  referred  to 
rectangular  axes. 

Let  A  be  the  point  of  pro- 
jection, and  P  the  position  of 
the  body  at  the  end  of  a  time  t. 
Take  the  origin  of  coordinates 
at  A,  x  horizontal,  and  y  vertical.  Let  a  =  EAF.  The 
coordinates  of  P  being 


214  KINETICS.  [299-301.1 

.     AF=x,    FP  =  y, 
we  have, 

AE=  vt, 

AF=  vt  cos  a  =  x,    .....     (1) 

EF  '=  vtsiua, 


—  %g#  =  y   ...    (2) 
Eliminating  t  from  equations  (1)  and  (2)  gives 


go? 
y  =  x  tan  a 


which  is  the  required  equation. 

299.  To  find  the  range  AB.     Making  y  =  0  in  equa- 
tion (3),  we  have 

aj  =  0, 
and 

x  =  4rA  cos  a  sin  a  =  2A  sin  2a  =  AB. 

300.  To  find  the  Time  of  Flight.     Substitute  the 
value  of  AB,  given  in  the  preceding  article,  in  equation 
(1)  of  Article  298,  and  we  find 

4A  sin  a  _     v  sin  a 

V    —  ~r  —     M  • 

v  g 

301.  To  find  the    greatest    height  in  the  path.  — 
It  will  be  vertically  over  1\  the  middle  point  of  AB. 
From  Article  299  we  find 

AD  =  2A  cos  a  sin  a  ; 

which,  substituted  for  x  in  equation  (3)  of  Article  298, 
gives 

CD  =  h  sin8  a. 


[302-304.] 


PROJECTILES. 


215 


302.  Greatest  Range. — To  find  the  angle  of  elevation 
of  the  projectile  which  will  give  the  greatest  range. 

This  requires  that  the  value  of  A£,  in  Article  299, 
shall  be  a  maximum  ;  hence, 

sin  2a  =  1 
/.  a  =  45°. 

303.  Greatest  Height. — To  find  the  angle  of  elevation 
of  projection  which  shall  give  the  greatest  height.     This 
requires  that  the  value  of  CD,  Article  301,  shall  be  a 
maximum;  hence, 

sin2  a  =  1 
/.a  =  90°; 

that  is,  the  projection  must  be  vertical. 

304.  Equal  Ranges. — Since  the  sine  of  an  angle  equals 


E 


FIG.  141. 

the  sine  of  its  supplement,  the  value  of  AJB,  Article  299, 
will  be  the  same 

for  sin  2a,  as  for  sin  (180°  —  2a) ; 
hence,  the  range  will  be  the  same  for  the  angles 

a,  and  90°  — a; 

that  is,  if  the  angles  of  elevation  of  two  projectiles  be  the 
complements  of  each  other,  and  have  the  same  initial 
velocity,  their  ranges  will  be  equal  to  each  other. 


216  KINETICS.  [305,306.] 

305.  Range  on  an  Inclined  Plane. — If  an  inclined 
plane,  OJB,  passes  through  the  point  of  projection,  it  is 
required  to  find  the  range  on  the  plane. 

Let  the  angle  BO  C=  <£,  then  will 

CB  =  y  =  x  tan  <j>, 

and  this  value  of  y,  substituted  in  equation  (3)  of  Article 
298,  gives 

2-y2  cos  a  sin  (a  —  d>) 


OB  =  UU  = 
therefore, 

OB=  6><7secd>  = 

g  cos**  <f) 

When  the  projectile  is  moving  in  a  vacuum,  gravity  is  the 
only  force  which  acts  upon  it. 

306.  Problems. — 1.    find  the  equation  of  the  path 
when  the  body  is  projected  horizontally. 

In  this  case  a  =  0  in  equation  (3)  of 
Article  298 ;  hence  we  have 


If  y  be  taken  positive  downwards,  we 
have 


FID.  142. 


2.  When  the  lody  is  projected  horizontally,  find  the 
range  B  C  on  a  horizontal  plane  below  the  point  of  pro- 
jection, and  the  time  offiight. 

Let  AB  =  yv  Substitute  y^  for  y  in  the  preceding 
equation,  and  solve  for  x,  and  we  find 


The  time  of  flight  will  be  that  necessary  to  move  the 


[306.]  PROJECTILES.  217 

horizontal  distance  B  C  with  the  velocity  v.    Let  t  be  the 
time,  then 


3.  To  find  the  velocity  and  angle  of  elevation  so  that  a 
projectile  shall  pass  through  two  points  whose  coordinates 
are  known. 

Let  #!  and  yt  be  the  coordinates  of  one  point,  0%  and  y2 
the  coordinates  of  the  other,  and  these  substituted  for 
x  and  y  respectively  in  equation  (3)  of  Article  298,  give 


4/1  cos"*  a 

x? 

/*  2    . 
4A  cos2  a 


Eliminating  h  gives 

i 
tan  a  =  ^ 


Eliminating  tan  a,  gives 
4 

hence, 


2  cos  a 


EXAMPLES. 


1.  A  body  is  projected  with  a  velocity  of  5g  at  an  angle  of 
elevation  of  45°  to  the  horizon ;  determine  the 
range  and  greatest  height. 


KINETICS.  [300.) 

2.  A  oody  is  projected  horizontally  with  a  velocity  equal 

to  that  acquired  by  a  body  falling  through  a  height 
of  15  feet;  required  the  range  J2G\  Fig.  142,  on  a 
plane  12  feet  below  the  point  of  projection. 

3.  The  horizontal  range  of  a  projectile  is  1,000  feet,  and 

,  time  of  flight  15  seconds ;  required  the  angle  of 
elevation,  velocity  of  projection,  arid  greatest  alti- 
tude. 

Ana.  a  =74°  33'  9". 

v  =  250.29  feet. 

A  =  904.69  feet. 

4.  A  body  projected  at  an  angle  of  elevation  of  45°  has  a 

horizontal  range  of  25,000  feet ;  required  the  velocity 
of  projection,  the  greatest  altitude,  and  time  of  flight. 
Ans.  v  =     896  feet. 
A  =  6,250  feet. 
t  =  39  +  seconds. 

5.  The  horizontal  range  of  a  projectile  is  four  times  its 

greatest  height ;  required  the  angle  of  elevation. 

6.  A  body  is  projected  from  the  top  of  a  tower  150  feet 

high,  at  an  angle  of  elevation  of  45°,  with  a  velocity 
of  75  feet  per  second  ;  required  the  range  on  the 
horizontal  plane  passing  through  its  foot. 

7.  The  eaves  of  a  house  are  25  feet  from  the  ground,  and 

the  roof  is  inclined  at  an  angle  of  30°  to  the  horizon, 
the  ridge  being  32  feet  from  the  ground ;  find  where 
a  smooth  sphere  sliding  down  the  roof  will  strike  the 
ground. 

8.  A  body  passes  through  two  points  whose  coordinates 

are  sst  =  400  feet,  y±  =•  60  feet,  x%  =•  600  feet,  and 
y2  =  50  feet ;  find  the  velocity,  angle  of  elevation, 
horizontal  range,  and  time  of  flight. 


|806.]  PROJECTILES.  219 

The  effect  of  the  resistance  of  the  air  may  be  illus- 
trated by  the  fact  that  a  certain  ball  being  pro- 
jected with  a  velocity  of  1,000  feet  per  second  in  the 
air,  its  range  was  found  to  be  about  5,000  feet,  instead 
of  31,250  feet,  as  it  would  have  been  in  a  vacuum. 

EXERCISES. 

1.  A  body  is  projected  vertically  upwards ;  what  will  be  its  range  ? 

2.  A  ship  is  sailing  due  east  at  the  rate  of  5  miles  per  hour ;  if  two 

bodies  are  projected  from  the  ship  at  the  same  angle  of  elevation, 
one  in  an  easterly  direction  and  the  other  in  a  westerly  direction, 
will  the  range  of  the  projectiles  be  the  same  ? 

3.  If  a  ship  sails  at  the  rate  of  15  miles  per  hour,  and  a  body  be  thrown 

from  it  in  an  opposite  direction  with  a  velocity  of  11  feet  per 
second,  required  the  actual  velocity  of  projection. 

4.  Considering  the  rotation  of  the  earth,  will  the  range  of  a  projectile 

be  the  same  if  it  be  fired  in  a  westerly  direction  that  it  will  if 
fired  in  an  easterly  direction  ? 

5.  If  a  projectile  be  struck  horizontally,  when  at  its  highest  point,  by 

another  body,  will  it  reach  the  horizontal  plane  in  the  same  time 
as  if  it  had  not  been  struck  ? 

6.  If  two  particles  be  projected  from  the  same  point  with  the  same 

velocity,  but  the  angle  of  elevation  of  one  is  as  much  above  45°  as 
the  other  is  below  it,  will  their  ranges  on  a  horizontal  plane  be  the 
same? 

7.  If  two  bodies  are  projected  from  the  same  point  on  a  smooth  hori- 

zontal plane,  with  different  velocities,  and  in  different  directions, 
show  that  the  straight  line  which  joins  them  alwayi  moves  paral- 
lel to  Itself. 


CHAPTER  XVI. 

CENTRAL   FORCES. 

307.  A  Central  Force  is  one  which  acts  directly  to- 
wards or  from  a  point.     The  point  is  called  the  centre  of 
the  force.     The  body  may  move  directly  towards  or  from 
the  centre,  or  it  may  move  about  it  in  a  curved  path,  called 
the  orbit.     The  latter  is  the  one  commonly  understood 
when  central  forces  are  referred  to.     Thus,  the  centre  of 
the  sun  is  considered  as  the  centre  of  the  attractive  force 
which  acts  upon  the  planets,  and  the  planets  move  in  orbits 
about,  the  sun.     Similarly,  the  centre  of  the  earth  is  con- 
sidered as  the  centre  of  the  force  which  acts  upon  the 
moon,  and  causes  it  to  move  in  its  orbit.     When  a  stone  is 
whirled  in  a  sling,  the  centre  of  the  force  which  continu- 
ally pulls  upon  the  body  is  at  the  point  where  the  string 
is  held  by  the  hand. 

[In  the  case  of  the  sun  and  planets,  the  centre  about  which 
they  revolve  is  not  the  centre  of  the  sun,  but  a  point  which  is  at 
the  common  centre  of  the  system,  and  which  is  relatively  near 
the  centre  of  the  sun.]  • 

308.  Centripetal  Force  is  a  name  given  to  that  central 
force  which  acts  towards  the  centre.     It  is  either  attract- 
ive, or  of  the  nature  of  a  pull.     Thus,  the  attractive  force 
of  the  sun  upon  the  planets,  and  of  the  primary  planets 
upon  their  secondaries,  is  centripetal.     The  pull  of  the 
string  upon  the  body  whirled  in  a  sling   is  centripetal. 
The   cohesive   force  of  the   metal  in  a  fly-wheel,  which 


[309,310.] 


CENTRAL   FORCES. 


221 


prevents  the  rim  from  flying  away  from  the  centre,  is  of 
the  same  nature. 

309.  When  a  body  moves  in  an  orbit,  the  centripetal 
force  constantly  draws  the  body  away  from  the  straight 
line  in  which  it  tends  to  move.     According  to  the  FIRST 
LAW  of  motion,  the  body  would  move  in  a  straight  line  if 
it  were  not  acted  upon  by  any  force,  but  by  the  constant 
action  of  the  central  force,  it  moves  in  a  curved  line.     The 
straight  line  in  which  the  body  tends  to  move,  is  a  tangent 
to  the  orbit,  as  DG,  Fig.  143. 

310.  The  Orbit. — If  a  body  at  rest  be  left  to  the  action 
of  a  central  force,  it  will  move  directly  towards  or  from 
that  centre.     The  same  will  be  true  if  it  be  projected 
directly  towards  or  from  the  centre.     In  such  cases  the 
orbit  is  a  straight  line  passing  through  the  centre  of  the 
force. 

Thus,  if  there  were  a  hole  through  the  earth  so  that  a 
body  could  move  from  surface  to  surface,  a  body  placed  in 
the  hole,  or  thrown  direct- 
ly  towards  or  from   the 
centre  of  the  earth,  would 
move  under  the  action  of 
a  central  force,  and   the 
path  would  be  a  straight 
line. 

But  if  the  body  be  pro- 
jected at  an  angle  with 
the  line  joining  the  body  Plo.  143- 

and  the  centre  of  force, 

it  will  move  in  a  curved  path.  The  orbits  of  aL  tn« 
planets  are  ellipses,  with  the  sun  at  one  of  the  foci. 
Thus,  if  ADB  be  an  ellipse,  representing  an  orbit  of  a 


G 


222  KINETICS.  [311,312.1 

planet,  C  and  C'  the  foci,  then  will  the  sun  be  at  one  of 
the  latter  points  in  reference  to  the  orbit.  It  is  generally 
assumed  that  comets  move  in  parabolic  orbits,  the  sun  be- 
ing at  the  focus.  This  is  done  so  as  to  simplify  the  com- 
putations for  determining  their  positions. 

311.  To  find  the  orbit  it  is  necessary  to  know  the  law 
of  action  of  the  central  force,  and  the  position,  velocity, 
and  direction  of  motion  of  the  body  at  some  point  in  the 
orbit.     In  the  solar  system  the  force  varies  inversely  as 
the  square  of  the  distance  from  the  centre.     See  Article 
65.     The  complete  investigation  of  this  subject  properly 
belongs  to  higher  mathematics,  and  we  shall  consider,  in 
this  chapter,  only  that  case  in  which  the  orbit  is  a  circle. 

312.  Centrifugal  Force. — This  is  a  name  given  to  a 
force  which  is  equal  and  opposite  to  that  which  deflects 
the  body  from  a  tangent  to  the  curve.     If  the  body  moves 
in  the  arc  of  a  circle  it  will  be  equal  and  opposite  to  the 
centripetal  force.     If  the  orbit  is  not  a  circle  the  central 
force  will  act  at  an  oblique  angle  to  the  tangent  at  nearly 
every  point  of  the  path.      In  Fig.  143,  let  the  body  be 
at  D,  and  the  centre  of  the  force  at  C.     Let  DF  repre- 
sent  the  magnitude  of  the  fo^ce,  and  let  it  be  resolved 
into  two  forces,  one,   EF,   parallel   to  the  tangent,   the 
other,  DE,  perpendicular  to  the  tangent.     If  the  body  is 
moving  in  the  direction  D6r,  the  component  EF  will  re- 
tard the  motion  ;  if  in  the  opposite  direction,  the  motion 
will  be  accelerated ;  but  in  either  case  the  component  DE 
pulls  the  body  directly  away  from  the  tangent.     Then  will 
the  centrifugal  force  be  equal  and  opposite  to  the  normal 
component  DE. 

[The  idea  is  often  conveyed  that  the  centrifugal  and  deflecting 
forces  act  upon  the  moving  body  at  the  same  time,  but  such  ia 


[313.]  MOTION  IN  A  CIRCLE.  223 

not  the  case  ;  for,  if  they  did,  they  would  exactly  neutralize  each 
other,  and  the  body  would  move  in  a  straight  line.  Thus,  when 
a  train  of  cars  runs  around  a  curve,  the  forces  under  which  the 
body  moves  are  the  propelling  force  of  the  steam  and  the  deflect- 
ing force  of  the  track.  The  deflecting  force  is  centripetal.  The 
so-called  centrifugal  force  is  exactly  equal  and  opposite,  and  is  a 
measure,  in  the  contrary  direction,  of  the  pressure  exerted  by  the 
rails  of  the  track.  If  we  consider  the  deflecting  force  as  acting 
between  the  rail  and  wheel,  then  will  the  action  of  the  force  upon 
the  wheel  be  centripetal,  and  the  reaction,  or  pressure  outward 
against  the  rail,  be  centrifugal.  Some  have  also  stated  that  the 
centrifugal  force  is  that  which  causes  the  body  to  fly  away,  or 
to  tend  to  fly  away,  from  the  centre,  as  if  there  were  an  active 
force  radially  outward,  due  to  the  revolution.  But  there  really 
is  no  such  force  in  this  case.  The  body  tends  to  go  in  a  straight 
line,  tangent  to  its  path.  Cut  the  string  of  a  sling,  or  let  go  of 
one  string,  and  the  body  starts  off  in  a  straight  line.  The  term 
centrifugal  is  not  so  objectionable  as  the  idea  which  it  has  been 
made  to  represent.] 


Motion  in  the  Circumference  of  a  Circle. 

313.  To  find  the  value  of  the  central  force  when 
the  body  describes  the  arc  of  a  circle.  Conceive  that  the 
body  is  held  by  a  string  fastened  to  a  fixed  point  at  the 
centre  of  the  circle ;  it  is  proposed  to  find  the  tension  of 
the  string  as  the  body  moves  along  the  circumference. 
To  solve  this  problem,  first  consider  the  conditions  by 
which  a  body  may  be  made  to  move  over  the  successive 
sides  of  an  inscribed  regular  polygon  with  a  uniform 
velocity.  Let  ABCD,  etc.,  be  a  regular  inscribed  poly- 
gon. If  a  particle  moves  over  the  side  AB  with  a  uni- 
form velocity,  it  is  required  to  determine  the  direction 
and  magnitude  of  an  impulse  applied  at  £,  that  shall 
cause  the  body  to  move  along  the  side  BC  with  the  same 
velocity  that  it  did  along  AB. 

Since  the  velocity  is  to  be  uniform,  the  sides  AB  and 


224 


KINETICS. 


[313.J 


BC  will    be   pro}X>rtional   to   the  velocities  along  those 
sides;  and  the  velocity  along  the  chord  BC  will  be  the 

resultant  of  that  along  AB,  and 
of  that  produced  by  the  re- 
quired impulse.  On  AB  and 
BC,  as  sides,  construct  the 
rhombus  ABCF\  then  will 
the  diagonal  BF  be  propor- 
tional to  the  velocity  which 
must  be  produced  by  the  re- 
quired impulse,  and  will  also 
represent  the  direction  in  which 
the  impulse  must  act.  See  Ar- 
ticles 14  and  15. 


FIG.  144. 


Let 


I  =  AB  =  BC  =  the  length  of  a  side  of  the  poly- 
gon; 

*  =  BF',  r  =  OB  =  the  radius  of  the  circle ; 
t  =  the  time  of  moving  over  a  side  of  the  polygon  ; 
v  =  the  velocity  with  which  the  particle  moves  ; 
m  =  the  mass  of  the  particle  ; 
Q  =  the  value  of  the  impulse ;  and 
V-L  =  the  velocity  imparted  by  the  impulse. 
The  diagonal  BF  bisects  the  angle  ABC  and  passes 
through  the  centre  of  the  circle.     Prolong  it  to  G,  and 
draw  the  chords  £67and  AC.     The  similar  right  angled 
triangles  BOG  and  £  EG  give 

BE  _BC 
•  BC~  BG' 
or, 


[313.]  CIECULAE  MOTION. 


Multiply  both  sides  of  this  equation  by  m  and  divide  by 
2,  and  it  may  be  written 


s 

m  - 


which  gives 


According  to  Article  122,  the  first  member  of  this  equa- 
tion is  the  force  which,  acting  during  a  time  t  (the  time  of 
moving  over  one  side  of  the  polygon),  will  produce  the  re- 
quired velocity  v^  ;  that  is,  we  have 


But,  in  order  that  the  motion  shall  be  along  the  straight 
line  BC,  the  velocity  v±  must  be  produced  in  an  instant. 
Let  F'  be  the  force  which  will  produce  the  velocity  vl  in 
an  element  of  time  At  ;  then,  according  to  Articles  122 
and  124,  we  have 

F'  =  ^  =  H.  (3) 

At       At 

If  this  force  acts  upon  the  particle  at  all  the  angles  of 
the  polygon,  the  particle  will  describe  the  successive  sides 
of  the  polygon  with  a  uniform  velocity. 

The  second  member  of  equation  (1)  is  independent  of 
the  number  of  sides  of  the  polygon,  and,  hence,  for  the 
same  circle  and  a  constant  velocity,  the  ratio  of  «f  to  t  in 
the  first  member  will  remain  constant.  If  the  number  of 


226  KINETICS.  [313.J 

sides  of  the  polygon  be  increased,  the  velocity  remaining 
constant,  the  time  of  describing  a  side  will  be  diminished, 
hence,  vi}  equations  (2)  and  (3),  will  decrease,  and  F'  ,  in 
equation  (3),  will  also  decrease.  Let  the  number  cf  sides 
of  the  polygon  be  increased  until  the  element  of  time  *lt, 
during  which  F'  acts,  is  the  same  as  that  occupied  by  the 
particle  in  describing  one  of  the  sides  of  the  polygon,  and 
let  Avi  be  the  velocity  which  it  will  produce  in  that  time; 
then  will  the  value  of  F'  be 

Til  AVl 

f    =  m  •=?  : 
At 

and  since  At,  Av^  and  v  are  simultaneous  quantities,  equa- 
tion (1)  becomes 

F'  =  m<b  =  mt.  (4) 

At  r 

Under  these  conditions  the  particle  describes  the  sides 
of  a  polygon  of  an  indefinitely  large  number  of  sides  with 
a  uniform  velocity.  But  the  limit  of  the  polygons  is  the 

^j?) 

circle  ;  the  limit  of  the  fraction  —  -1  is,  according  to  equa- 

At 

v* 
tion  (4),  the  constant  quantity  —  ;  and  the  limit  of  F'  is 

some  constant,  whose  value  equals  the  limiting  value  of 
the  second  member  of  equation  (4).  But  what  is  con- 
stantly true  of  a  value  as  it  approaches  a  limit  indefin- 
itely* is  true  of  the  limit  /  hence,  calling  f  the  limiting 
value  of  F',  we  have,  for  the  value  of  f^  ^ohen  the  motion 
is  in  the  arc  of  a  circle, 


which  is  the  constant  force  which  acts  towards  the  centre 


[314-S16,]  CENTRIFUGAL  FORCE.  227 

It  follows  from  this  equation  that  the  deflecting  force 
varies  directly  as  the  mass  of  the  body  and  the  square  of 
its  velocity,  and  inversely  as  the  radius  of  the  circle. 

314.  To  find  the  value  of  the  central  force  in  terms  of 
the  Angular  Velocity. 

Let  o>  =  the  angular  velocity,  then 

v  =  ro>  ; 
and  equation  (5)  becomes 

f=  mra?. 

315.  To  find  the  value  of  the  Centripetal  Force  in  terms 
of  the  time  of  a  complete  revolution. 

The  time  of  a  complete  revolution  is  called  the  Periodic 
time  of  the  motion. 

Let  T=  the  periodic  time  ;  then 


27TT 

.-.  v  =  ~ 


which,  substituted  in  equation  (5)  of  Article  313,  gives 


316.  Centrifugal  Force  of  Bodies  of  finite  size.  — 

Let  T/&J,  TM.J,  etc.,  be  the  masses  of  the  particles  of  the  body, 
*"ij  Tzi  etc-,  be  their  respective  distances  from  the  axis  of 
revolution,  M  the  total  mass  of  the  body,  and  r  the  point 
at  which,  if  the  whole  mass  of  the  body  were  concentrated, 
the  centrifugal  force  would  remain  the  same  ;  then  we 
have 

.)o>  =  aJtmr; 


228  KINETICS.  [317,318.] 

hence,  according  10  Article  216,  the  centrifugal  force  will 
be  the  same  as  if  the  whole  mass  of  the  body  were  concen- 
trated at  its  centre  of  gravity  and  1'evolved  with  the  sam<, 
angular  velocity. 

Problems. 

317.  Find  the  centrifugal  force  on  the  equator  due  to 

the  revolution  of  the  earth  on  its  axis. 

The  time  of  the  revolution  of  the  earth  on  its  axis  ia 
T  =  86,164  seconds  of  mean  solar  time  ;*  the  equatorial 
radius  of  the  earth  is  20,923,161  feet  ;f  and  TT  =  3.14159. 
These  values  in  the  equation  of  Article  315  give 

/=  0.1112m. 

The  force  of  gravity  exceeds  this  value  ;  for  it.  is  found 
that  a  body  on  the  equator  weighs  (see  page  32), 

W=  mg  =  32.0902m  ; 

hence,  if  the  earth  did  not  rotate,  and  other  things  re- 
mained as  at  present,  a  body  would  weigh 

W1  =  W+f=  32.2014m. 
We  also  have 

f          0.1112m        J_ 

If;  ~  32.2014m  ~  289  n 

hence,  the  rotation  of  the  earth  causes  a  diminution  of  the 
weight  of  bodies ;  the  diminution  on  the  equator  being  ¥|^ 
of  their  original  weight. 

318.  In  what  time  must  the  earth  revolve  so  that  bodie* 
on  the  equator  will  weigh  nothing  f 

*  The  earth  makes  a  complete  revolution  in  less  than  24  hours  mean 
time. 

f  See  foot  note  on  page  33. 


[319,320.]  PROBLEMS.  229 

Let  TI  be  the  required  time,  then  will/j  in  Article  315, 
equal  W^  aud  we  have 


and  dividing  the  equation  of  Article  315  by  this  equation, 
calling  T  the  periodic  time  of  the  revolution  of  the  earth, 
we  have 

f_   _  T?  _  _\_ 

W  ~~  T*~  289 


319.  To  find  the  central  (centripetal)  force  necessary  to 
keep  the  moon  in  its  orbit. 

The  mean  time  of  the  periodic  motion  of  the  moon  is 
T  =  2,360,585  seconds  ;  and  the  mean  distance  of  the 
moon  from  the  centre  of  the  earth  is  60.361  Jt,  in  which 
M  is  the  mean  radius  of  the  earth.  Calling  R  =  20,897,- 
500  feet  and  substituting  these  quantities  in  the  equation 
of  Article  315,  gives,  for  a  unit  of  mass, 
/=  0.0089. 

320.  Wliat  is  the  force  of  gravity  at  the  moon  ? 

The  force  of  gravity  varies  inversely  as  the  square  of  the 
distance  from  the  centre  of  the  earth.  (See  Article  65.) 
Calling  the  mean  value  of  g  at  the  surface  of  the  earth 
32.246,  we  have 


If  the  data  in  the  two  preceding  problems  were  correct 
in  every  particular,  the  results  should  be  exactly  alike, 
that  is,  /'  =  g'.  In  this  way  Sir  Isaac  Newton  tested  the 
law  of  UNIVERSAL  GRAVITATION. 


230 


KINETICS. 


[321,323.; 


321.  A  locomotive  whose  weight  is  W  tons  runs  around 
a  horizontal  curve  whose  radius  is  r  feet,  with  a  velocity 
of  V  miles  per  hour  ;  required  the  centrifugal  force  y 
that  is,  the  pressure  against  the  outer  rail. 

These  quantities,  reduced  to  the  units  of  equation  (5), 
Article  313,— that  is,  to  feet,  pounds,  and  seconds — and 
substituted  in  that  equation,  give 


J 


2,000  W 


/_5280_     \2 
\60  x  60      / 


32* 


k  W(in  tons)  x  V\in  m.  pr.  A.)  ,, 

=  130 r. — -±— — Ids.,  nearly. 

r  (in  jeet) 

322.  To  find  the  elevation  of  the  outer  rail  so  that 
the  resultant  pressure  of  a  locomotive,  as  it  passes  around. 
a  curve,  will  be  perpendicular  to  the  plane  of  the  track. 

Let  DE  represent  the  weight, 
W,  of  the  locomotive,  and  EFi\\e 
centrifugal  force  ;  then  will  DF 
be  the  resultant  of  these  forces. 
The  elevation  of  the  outer  rail 
must  be  such  that  DFw\\\  be  per- 
pendicular to  the  line  joining  the 
tops  of  the  rails,  which,  in  prac- 
tice, will  be  parallel  to  the  line  A  C.  Draw  AB  horizontal 
and  CB  vertical. 

Let  b  =  AB\  h  =  £C;  r  =  the  radius  of  the  curve  : 
and  v  =  the  velocity  of  the  locomotive. 

Then  the  similar  triangles  DEF  and  AB  C  give 

DE  \EF\\AB\BG\ 


A. 


or, 


W:f::b:h-, 


r  333.1 


PROBLEMS. 


231 


S  abstitute  the  value  of  f,  equation  (5),  Article  313,  and 
W=  mg,  and  we  hare 

v>   I 

fi  =  — .— , 

g  r 

in  which  v  and  g  are  both  measured  in  feet  per  second,  and 
b  and  r  in  feet.  In  practice,  the  elevation  is  usually  so 
small  that  b  is  taken  equal  to  A  C,  the  gauge  of  the  track. 
It  will  be  seen  that  the  elevation  is  independent  of  the 
weight  of  the  body,  and  that  it  varies  as  the  square  of  the 
velocity. 

323.  To  determine  the  motion  of  a  Conical  Pendu- 
lum.— A  conical  pendulum  is  a  heavy  body  revolving 
about  a  vertical  axis.  The  governor  of  a 
steam  engine  is  an  example.  The  forces 
which  act  upon  the  body  B  are  the  force 
of  gravity,  which  equals  the  weight  of  the 
body,  the  tension  of  the  piece  J3A,  and 
the  force  which  produces  the  rotation- 
Let  the  body  be  at  rest  and  a  force,  equal 
to  the  centrifugal  force,  act  upon  the 
body  ;  then  will  the  relation  of  the  parts 
be  the  same  as  before. 

Let  W  =  Bb,  and  resolve  it  into  two  forces,  one,  Be, 
horizontal ;  the  other,  be,  parallel  to  BA  ;  then  will  the 
former  represent  the  centrifugal  force,  and  the  latter  the 
tension  on  BA.  Let  m  =  the  mass  of  the  body  and  v  its 
velocity,  then,  according  to  Article  313, 

D  vz  v* 

±fc  =  m  -^7,  =  m    .  r>   -. — T. 
B  C          AB  sin  < 


FIG.  146. 


232  KINETICS.  [333.] 

We  also  have 

BG  =  W  tan  Bbc  -  Ftan  </> ; 

• — r  =  m<f  tan  9  ; 


Sill  i. 

from  which  we  find 

v  =  VAJS  .gsiiKJ)  tan  0. 
Let  T  =  the  time  of  one  revolution,  then 

T  = 


.  sn 


.  <?  sin  </>  tan 


—  cos 

v    g 

from  which  we  find 


hence,  the  angle  is  independent  of  the  mass  of  the  body. 

In  the  governor  the  balls  are  not  permitted  to  move  out 
freely,  but  are  required  to  overcome  a  resistance.  The 
resistance  may  be  reduced  to  an  equivalent  horizontal 
force  applied  at  B.  Let  F  be  this  force  ;  then  will 

/=  TFtari  0  +  F\ 

i? 
.:  F—  m  -       —  TFtan  <> 


=  W( TTT-T  -A  ~  tan  W 

I  g .  AB.  sin  <j>  T  J 


[328.]  PROBLEMS.  233 

EXAMPLES. 

1.  A  body  revolves  about  a  point  in  a  horizontal  plane  to 

which  it  is  attached  by  means  of  a  cord  ;  required 
the  velocity  of  the  body  so  that  the  tension  on  the 
cord  shall  be  twice  the  weight  of  the  body. 

2.  In  the  preceding  example,  if  the  radius  of  the  circle  is 

2  yards,  what  must  be  the  number  of  revolutions  per 
minute  so  that  the  tension  on  the  string  will  be  three 
times  the  weight  of  the  body  ? 

3.  A.  body  is  attached  to  a  point  by  means  of  a  cord  2  feet 

long,  and  revolves  uniformly  in  a  vertical  circle  ;  re- 
quired the  number  of  revolutions  per  minute  so  that 
the  tension  of  the  cord  shall  be  zero  when  the  body 
is  at  the  highest  point  of  the  circle. 

4.  In  the  preceding  example,  what  will  be  the  tension  of 

the  cord  when  the  body  is  at  the  lowest  point  of  the 
circle  ? 

5.  A  body  is  placed  on  the  inside  of  the  rim  of  a  wheel 

which  revolves  about  a  vertical  axis  ;  if  the  weight 
of  the  body  is  TF,  the  radius  of  the  wheel  7?,  and  the 
coefficient  of  friction  /*,  what  must  be  the  number  of 
revolutions  of  the  wheel  per  minute  so  that  the  fric- 
tion due  to  the  centrifugal  force  will  just  prevent 
the  body  from  sliding  vertically  downward  ? 

6.  A  body  is  placed  in  a  groove  in  a  horizontal  disc,  and 

attached  to  the  centre  of  the  disc  by  means  of  a 
string  30  inches  long.  The  disc  is  revolved  about  a 
vertical  axis  through  its  centre  at  the  rate  of  250 
revolutions  per  minute ;  the  coefficient  of  friction 
between  the  body  and  disc  being  0.15,  what  will  be 
the  tension  of  the  string,  the  groove  being  radial. 


234:  KINETICS.  [323.J 

7.  A  car  runs  around  a  curve  whose  radius  is  2,500  feet ; 

if  the  rails  are  in  a  horizontal  plane,  and  a  body  rests 
on  the  floor  of  the  car  between  which  and  the  body 
the  coefficient  of  friction  is  0.10,  what  must  be  the 
velocity  of  the  car  in  miles  per  hour  so  that  the  body 
will  be  just  on  the  point  of  sliding  outward  ? 

8.  What  must  be  the  elevation  of  the  outer  rail  so  that  a 

car,  moving  on  a  curve  whose  radius  is  3,000  feet, 
with  a  velocity  of  30  miles  per  hour,  shall  press 
equally  upon  both  rails,  the  distance  between  the  rails 
being  4  feet  8  inches. 

9.  A  weight  is  suspended  from  a  point  in  the  roof  of  a 

car  by  means  of  a  string  6  feet  long ;  the  car  runs 
around  a  curve,  whose  radius  is  4,000  feet,  at  the 
rate  of  40  miles  per  hour  ;  how  much  will  the  string 
deviate  from  a  vertical  on  account  of  the  deflecting 
force  ? 

10.  In  Fig.  14^5,  if  A  B  is  15  inches,  and  the  body  makes 

100  revolutions  per  minute,  find  the  angle  BA  C\ 
and  the  distances  AC  and  J3C. 

11.  In  Fig.  146,  if  the  resistance  which  the  centrifugal 

force  has  to  overcome  is  4  Ibs.,  acting  horizontally 
when  reduced  to  the  point  B,  the  weight  of  B  5  Ibs., 
and  the  length  AB,  14  inches;  what  must  be  the 
number  of  revolutions  per  minute  so  that  the  resist- 
ance will  be  overcome  when  the  angle  13 AC  is 
1 0  degrees  ? 

12.  If  a  grindstone  whose  diameter  is  4  feet,  thickness  -i 

inches,  tenacity  600  pounds  per  square  inch,  revolves 
about  an  axis  through  its  centre,  how  many  revolu- 
tions must  it  make  per  minute  to  produce  rupture 
along  a  diameter,  no  allowance  being  made  for 
the  eye. 


[323.]  EXERCISES.  235 


EXEKCI8E8. 

1.  If  a  body  moves  in  a  curved  line  which  is  not  the  arc  of  a  circle 

tinder  the  action  of  a  central  force,  will  the  deviating  force  be  the 
same  as  the  central  force  ? 

2.  Why  will  the  water  in  a  rotating  vessel  be  highest  around  the  out- 

side of  the  vessel  ?     Will  this  be  true  of  anything  besides  water  ; 
as  grain,  or  pebbles  ? 

3.  If  the  rotation  of  the  earth  were  to  cease,  about  how  much  would 

the  water  in  the  ocean  be  raised  at  the  poles,  and  how  much  would 
it  bo  depresso.l  afc  the  equator,  the  earth  being  considered  as  fluid  ? 

4.  Does  the  centrifugal  force  have  any  effect  upon  bodies  or  particles 

below  the  surface  of  the  earth,  as  in  a  deep  mine,  for  instance  ? 

5.  If  the  earth  were  a  hollow  sphere  and  water  were  thrown  into  the 

hollow,  where  would  it  come  to  rest  ? 

6.  If  the  earth  were  to  revolve  on  its  axis  once  in  84  minutes,  what 

would  happen  to  bodies  on  the  equator  ?     Would  the  cohesion  of 
the  parts  prevent  their  being  thrown  off  ? 

7.  If  the  moon  retained  a  circular  orbit,  but  should  revolve  around  the 

earth  every  15  days,  would  it  be  nearer  or  more  remote  from  the 
earth  than  at  present  ? 

8.  Why  do  those  planets  near  the  sun  go  around  it  in  less  time  thai 

those  more  remote  ? 

9.  Does  elevating  the  outer  rail  destroy  the  centrifugal  force  of  a  mov- 

ing train  ? 

10.  Water  is  put  on  the  face  of  a  grindstone  and  the  stone  revolved  so 

rapidly  that  the  water  flies  off  ;  does  it  go  off  radially  or  tan- 
gentially  ? 

11.  Do  the  centripetal  or  centrifugal  forces  have  anything  to  do  with 

the  velocity  with  which  the  stone  leaves  the  sling  ? 

12.  Clothes  may  be  partially  dried  by  placing  them  within  a  perforated 

vessel  which  is  made  to  revolve  very  rapidly  ;  explain  the  principle. 

13.  Are  bodies  at  the  poles  of  the  earth  affected  in  their  weight  on  ac- 

count of  the  rotation  of  the  earth  ?     Show  why  they  weigh  more 
there  than  they  would  if  the  earth  ceased  to  rotate  ? 

14.  If  a  train  of  cars  runs  around  a  circular  track  in  which  both  rails 

are  in  the  same  horizontal  plane,  is  there  any  danger  of  the  cars 
being  overturned  by  the  centrifugal  force  ? 


CHAPTER  XYII. 

SOLUTION  OF  PROBLEMS  IN  WHICH  THE  INTENSITY  OF  THB 
FORCE  VARIES  DIRECTLY  AS  THE  DISTANCE  FROM  THF 
CENTRE  OF  THE  FORCE,  AND  IS  ATTRACTIVE. 

324.  "When  a  perfectly  elastic  solid  is  pulled  or  com- 
pressed, or  distorted  in  any  manner,  the  force  which  resists 
distortion  varies  directly  as  the  amount  of  distortion,  if  the 
elastic  limits  are  not  exceeded.     See  Article  131.     This 
law,  in  which  the  force  varies  directly  as  the  distance, 
holds  good  in  several  other  problems. 

General  Formulas. 

325.  To  find  the  Velocity. — If  a  body  starts  from  rest 
and  is  acted  upon  by  a  force  which  varies  directly  as  the 
distance  from  the  centre  of  the  force  ;  required  the  velo- 
city of  the  body  when  it  reaches  the  centre  of  the  force, 
and  also  the  velocity  at  any  point  of  the  path. 

Let  A,  Fig.  147,  be  the  origin  of  the  force,  B  the  point 
^  where  the  particle  starts,  and  the 
line  AB  the  path  along  which  the 
particle  moves.  Erect  BC  to  re- 
present the  intensity  of  the  force 
at  _Z>,  and  draw  the  straight  line 


6  AC\    then  will  any  ordinate    be 

Fio.  147.  Al         P  ,. 

represent   the  force  acting    upon 

the  particle  when  it  is  at  b.     The  body,  starting  from  rest 
at  JS,  will  be  constantly  accelerated  until  it  arrives  at  A, 


|325.1 


GENERAL  FORMULAS. 


237 


but  the  rate  of  acceleration  will  continually  decrease  from 
£  to  A,  because  the  force  decreases  in  intensity. 

To  represent  the  velocity  approximately  by  a  geometri- 
cal construction,  divide  the  space  Alt,  Fig.  148,  into  equal 
small  spaces,  Bf,  fg,  gh,  etc.,  and  erect  ordinates,  fa,  gc, 
he,  etc.  Consider  the  force  as  constant  while  the  particle 
is  passing  from  JS  tof,  and  represented  by  the  arithmeti- 
cal mean  between  JBC  and  fa.  This  force  will  produce  a 
certain  velocity,  which  represent  by  Bm.  Draw  mn  par- 
allel to  AB,  and  at  n,  where  it  intersects  af  prolonged, 


— 'at 


o 


FIG.  148. 


draw  no  to  represent  the  velocity  produced  by  \(af  +  cg\ 
and  draw  op  parallel  to  AJ3,  to  meet  eg  prolonged  at  j>, 
and  so  on.  The  points  B,  n,  p,  etc.,  will  be  the  vertices 
of  a  polygon  ;  and  by  reducing  the  spaces  Bf,  fg,  etc.,  in- 
definitely, the  polygon  will  approach  a  curve,  BD,  as  a 
limit.  Any  ordinate  to  this  curve,  as  bd,  Fig.  149,  will 
represent  the  velocity  of  the  particle  when  it  has  reached 
the  point  b  of  its  path. 

To  find  a  formula  for  the  velocity,  let 

t\  =  the  intensity  of  the  force  at  a  unit's  distance 
from  the  centre  of  the  force ; 

«o  =  AB  =  the  distance  of  the  particle  from  A 
when  motion  begins ; 


238  KINETICS.  [325.] 

s  =  Ab  =  any  distance  from  A  ; 

v  =  the  velocity  of  the  particle  at  b  ; 

Vi  =  the  velocity  of  the  particle  at  A  ;  and 

m  =  the  mass  of  the  particle  ; 
then  will 

i}80  =  BC  =  the  intensity  of  the  force  at  B\  and 

rjs  =  be  =  the  intensity  of  the  force  at  b. 
The  work  done  by  the  force  upon  the  particle  while 
moving  it  from  B  to  A,  will  be  represented  by  the  trian- 
gle ABC,  see  Article  96,  and  hence,  will  be 

i«0-^o  =  iW; 

which  will  impart  to  the  body  an  amount  of  kinetic  energy 
expressed  by 

8 

and 

*WI*5        ...    (1) 

hence,  the  velocity  at  the  centre  of  the  force  will  vary 
directly  as  the  distance  over  which  the  body  moves,  and 
directly  as  the  square  root  of  the  intensity  of  the  force  at 
a  unit's  distance  from  the  centre  of  the  force. 

The  work  done  while  moving  from  b  to  A  will  be 


hence,  the  work  done  in  passing  from  B  to  b  will  be  the 
difference  of  these,  or, 


.-.  v  =  V  —  *V  -  «*,         •        •    (2) 

Yfli 

which  gives  the  velocity  at  any  point  of  the  path. 


[326.]  GENERAL  FORMULAS. 

The  last  equation  may  be  written 

mv*  +  yd*  =  r)*02 ;   .        .        .    (3) 

and  since  v  and  s  are  variables,  and  all  the  other  quanti- 
ties constants,  it  is  the  equation  of  an  ellipse.  Hence,  the 
curve  BDE,  Fig.  149,  is  an  ellipse,  of  which  the  semi 
axis  AB  is  s0,  and  that  of  AD  is  found  by  making  s  =  o, 
and  finding  the  value  of  v.  This  value  of  v  becomes  vit 
and  is 


m 

as  given  in  equation  (1). 

After  the  particle  arrives  at  A  it  will,  by  virtue  of  the 
kinetic  energy  of  the  body,  pass  that  point  and  move  on 
until  the  force  at  A  overcomes  the  energy,  when  it  will 
stop  and  return.  The  distance  AJZwill  equal  AB.  The 
entire  distance  BE  is  called  the  amplitude.  This  motion 
is  called  oscillatory  or  vibratory. 

326.  To  find  the  time  of  the  movement  of  a  particle 
from  any  distance  to  the  centre  of  the  force,  when  the 
force  varies  directly  as  the  distance  from  the  centre. 

Equation  (2)  of  the  preceding  article  may  be  written 

9  9  ' W       9 

sa 3  —  tr  =  —  v* ; 

which  may  be  represented  by  a  right-angled  triangle,  ill 
which  SQ  is  the  hypothenuse,  s  one  side,  and  y  —  v,  the  other 

side.  If  with  A  as  a  centre,  and  a  radius  AD  —  sn,  a 
quadrant  be  described,  then  will  BD,  the  sine  of  the  arc 
CD  to  the  radius  of  the  arc,  and  AB,  the  cosine  of  the 


240  KINETICS.  [326.] 

same  arc,  constantly  represent  the  relation  between  the 
space  *,  and  the  y  —  times  the  velocity  v. 

The  velocity  is  constantly  varying ;  hence,  at  any  instant3 
we  have,  according  to  Article  10, 

~r        As  (i  \ 

At  =  —>       -        •        •        C1) 
v 

in  which  ~^~s  is  the  increment  of  space  passed  over  in  the 
corresponding  increment  of  time.     In  Fig.  151,  let  AD 


=  *,  and  AB  =  4       v. 


On  the  line  DB,  take  Db  to  represent  As,  and  draw  ba 
parallel  to  AB  to  meet  the  tangent  Da  drawn  through  D. 


Fio.  150 


The  triangles  Dba  and  ABD  are  similar,  having  the  sides 
of  the  one  perpendicular  respectively  to  the  sides  of  the 
other.  Hence,  we  have 

AB\  AD \\Dl\Da\ 


or 


/m  

V  -  v  :  s0 : :  As  :  Da ; 


[326.J  GENERAL  FORMULAS.  241 

•    (2) 


m      v     • 
which,  combined  with  equation  (1),  gives 

Da=  V  m  s°'4i> 
hence, 


.        .    (3) 

If  Da  be  diminished  indefinitely,  it  will  approach  the 
arc  as  a  limit  ;  heuce,  an  element  of  the  time  equals  an 

element  of  the  arc  divided  by  \/—  SQ  :    and  the  time  of 

v  m 

passing  from    O  to  A.  will  equal  the  quadrant    CDE, 

divided  by  {/  ?L  st.     Let  4  be  the  required  time,  then 

m 

_          jfrTfr          i  V^ 

(A\ 


Hence,  this  remarkable  result,  that  the  time  of  the  move- 
ment of  a  particle  from  any  point  to  the  centre  of  the  force, 
when  the  force  'caries  directly  as  the  distance  from  the 
centre,  is  INDEPENDENT  OF  THE  DISTANCE,  The  times,  there- 
fore, are  isochronous. 

Hence,  also,  the  time  will  be  the  same  for  all  distances 
from  the  centre  •  and  for  the  same  body  it  will  vary  in- 
versely as  the  square  root  of  the  intensity  of  the  force  at 
a  unites  distance  from  the  centre  of  the  force. 

Let  t  =  the  time  of  moving  through  the  amplitude  ; 
then 

(5) 


KINETICS. 


[327.J 


Problems. 

327.  Simple  Pendulum.  —  A  simple  pendulum  is  a 
material  particle,  suspended  by  a  mathematical  line,  and 
swinging  under  the  action  of  gravity. 

Let  O  be  the  point  of  suspension,  P  the  position  of  the 
particle,  and  w  =  mg  the  weight  of  the  particle.     Resolve 
the  weight  w  into  two  components,  one,  ba,  parallel  to 
OP;   the  other,  Pb,  perpendicu- 
lar to  OP,  which  will  also  be  tan- 
gent to  the  arc  BA.     The  line 
OA    being    vertical,    the    angle 
baP  will  equal  that  at  O.     The 
component  ab  will  be  resisted  by 
the  tension  of  the  cord  OP,  and 
the  motion  will  be  produced  by 
the  component  bP.     We  have 
bP  =  w  sin  a, 
=.  mg  sin  O, 


AP 

™>g  -       nearly, 


Fio.  152. 


when  the  angle  O  is  small.  As  OP  is  a  constant  radius, 
it  follows  that,  for  small  angles  of  oscillations,  the  moving 
force  varies  directly  as  the  distance  AP,  of  the  particle 
from  the  lowest  point  A.  Hence,  the  time  of  an  oscilla- 
tion may  be  determined  from  equation  (5).  For  this  pur- 
pose we  must  find  the  value  of  rj.  "We  have 

AP 

bP         g~OP  _    mg  . 
77  ~  AP      ~TP          OP  ' 


which,  substituted  in  equation  (5),  and  making  OP  =  lt 
gives 


[328.]  PROBLEMS.  243 


•4-    __ 


which  is  the  value  given  in  Article  67.     In  practice  a 
compound  pendulum  is  always  used. 

328.  Compound  Pendulum. — Any  body  of  finite  size, 
vibrating  under  the  action  of  gravity,  is  a  compound  pen- 
dulum. It  is  shown,  both  by  analysis  and  by  experiment, 
that  there  are  always  two  points  in  a  compound  pen- 
dulum, about  which  the  body  will  oscillate  in  the 
same  time,  and  the  distance  between  these  points  is 
the  length  of  an  equivalent  simple  pendulum. 
Thus,  if  the  body  be  suspended  at  A,  and  the  num- 
ber of  vibrations  be  noted,  then  there  is  another 
point,  B,  at  which,  if  it  be  suspended,  it  will  vibrate 
in  the  same  time,  and  the  distance  AB  will  be  the  FIG.  153. 
length  of  the  equivalent  simple  pendulum.  The 
point  A  is  called  the  centre  of  suspension  and  B  the  cen- 
tre of  oscillation.  It  requires  very  accurate  measurements, 
and  very  close  observations  to  determine  the  length  of  the 
pendulum  which  oscillates  in  a  certain  time,  but  it  has  been 
found  many  times  in  different  places  and  the  results  have 
been  recorded.  Having  found  this  result,  the  length  of  a 
pendulum  which  will  oscillate  once  in  a  second  may  be 
found  as  follows : 

Let  19  =  the  length  of  the  pendulum  which  will  oscillate 
once  in  a  second  ;  then,  according  to  the  equation  of  the 
preceding  article,  we  have 

I  sec.  —  TT  \/-. 

[  g 

Dividing  this  equation  by  the  former  one,  and  solving 
for  ln  gives 


244 


KINETICS 


[329,  330.; 


From  the  preceding  equation  we  have 

9  =  *%  5 

by  means  of  which  the  acceleration  due  to  gravity  may  be 
found. 
329.  Length  of  the  Second's  Pendulum. 

TABLE  GIVING  THE  LENGTH  OF  THE  SECOND'S  PENDULUM  AI 
DIFFERENT  PLACES  ON  THE  EARTH,  AND  THE  ACCELER- 
ATION DUE  TO  GRAVITY  AT  THOSE  PLACES. 


Observer. 

Place. 

Latitude. 

Length  of 
seconds 
pendulum 
in  inches. 

Accelerating 
force  of  gra- 
vity ;  feet 
per  second. 

Sabine  

Spitzbergen  

N.  79  "50' 

39-21469 

32-2528 

Hammerfest  ... 

TOW 

39-19475 

32-2863 

Svanberg  

Stockholm  

59  21' 

39-16541 

32-2122 

Bessel    

Konigsberg    .  .  . 

54°42' 

39-15072 

32-2002 

Sabine   

Greenwich.  

51°29' 

39-13983 

32  1912 

Borda,  Biot,   and 

Paris  

48°50' 

39-12851 

32-1819 

Biot  

Bordeaux    . 

44°50' 

39-11296 

32  1691 

Sabine  

New  York  

40°43' 

39-10120 

32-1594 

Freycinet  

Sandwich  Islands 

20\52 

39-04690 

32-1148 

Sabine  
Freycinet    .... 

Trinidad  
fiawak. 

10W 
S.    0°  2 

39-01888 
39  01433 

32-0913 

32-0880 

Sabine    and    Du- 
perrey  

Ascension  

7°55' 

39-02363 

32  0956 

Freycinet      and 
Duperrey  
B  risbane  and 
Ruinker.  

Isle  of  France  .  .  . 
Paramatta         .  . 

20°10' 
33°49' 

39-04684 
39  •  07452 

32-1151 
32-1375 

Freycinet      and 
Duperrey  

Isles  Malouines  .  . 

51°35' 

39-13781 

32-1895 

330.  To  find  the  number  of  seconds  lost  by  a  clock 

when  carried  to  a  given  height  above  the  surface  of  t/u 
earth. 


[330.]  PROBLEMS.  2 

Let  the  clock  on  the  surface  of  the  earth  indicate  mean 
solar  time,  then  in  one  day  it  indicates  86,400  seconds; 
when  taken  to  a  height  above  the  earth,  the  vibrations  of 
the  pendulum  will  be  slower,  because  the  force  of  gravity 
will  be  less. 

Let  N '=  86,400,  N^  =  the  number  of  seconds  indicated 
by  the  clock  when  at  a  height  A,  t  =  the  time  of  one  vibra- 
tion on  the  surface,  ^  =  the  time  of  one  vibration  at  the 
height  A,  and  r  =  the  radius  of  the  earth.  The  length  of 
the  pendulum  remaining  the  same,  we  have 


But 


hence, 


which,  subtracted  from  .ZV",  gives 


The  quantity  N —  &[  is  called  the  rate  of  the  clock. 
The  quantities  r  and  A  must  be  of  the  same  denomination. 


246 


KINETICS. 


[331.] 


If  the  loss  in  a  day  be  known  we  may  find  the  height, 
for  we  find 


z  _ 


331.  To  find  the  time  in  -which  a  body  -would  pass 
through  the  earth  from  surface  to  surface,  if  it  could 
pass  freely  without  resistances,  the  earth  being  considered 
as  homogeneous. 

If  the  earth  were  a  homogeneous  sphere,  the  attractive 
force  would  vary  directly  as  the  distance  from  the  centre 
of  the  earth  ;  see  Article  78.    Hence, 
if  r  be  the  radius  of  the  earth,  we 
have 

mg 

17  =  —  ; 

r    ' 

which,  substituted   in   equation   (5), 
Article  326,  gives 


t  = 


and  in  equation  (1)  of  Article  325,  gives 


i.    -    (D 


Equation  (1),  compared  with  the  value  of  t  in  Article 
327,  shows  that  the  time  required  for  the  particle  to  pass 
through  the  earth  equals  the  time  of  one  oscillation, 
through  a  small  arc,  of  a  simple  pendulum  whose  length 
equals  the  radius  of  the  earth. 

Equation  (3)  of  Article  72  gives  v  =V2gh  ;  which,  com- 
pared with  equation  (2)  above,  shows  that  the  velocity  at 
the  centre  of  the  earth  will  equal  -|\/2  times  the  velocity 
acquired  by  a  body  falling  freely  through  a  distance  equal 
to  the  radius  of  the  earth,  under  the  action  of  a  constant 


[332.1  PROBLEMS.  247 

force  equal  to  the  force  of  gravity  at  the  surface^  of  the 
earth. 

332.  Vibration  of  an  Elastic  Bar.—  Let  AB  be  a 
prismatic  bar,  having  a  weight  P  suspended  at  its  lower 
end.     If  the  weight   be  pushed  up  or  down 
slightly,  or  be  struck,  or  in  any  other  manner 
be  disturbed  in  a  vertical  direction,  it  will  oscil- 
late up  and  down  ;  but,  in  the  case  of  a  solid 
bar,  the  oscillation  will  be  small.     The  longi- 
tudinal  vibrations  of    rubber,  coiled  springs, 
and  the  like,  may  readily  be  seen. 

Transverse  vibrations,  which  really  follow 
the  same  law,  may  readily  be  seen  in  the  case 
of  solid  bars,  as  may  be  illustrated  by  a  tuning 
fork. 

Suppose  that  AB  is  the  length  of  the  bar  when  P  is 
suspended  at  its  lower  end,  and  that  by  some  means  it  is 
elongated  to  b.  When  the  disturbing  force  is  removed 
the  elastic  force  in  the  bar  will  pull  the  weight  up  and  the 
end  will  pass  the  point  B  and  rise  to  some  point  as  c,  in 
which  condition  the  bar  will  be  in  a  state  of  compression. 
The  weight  will  then  descend  to  &,  and  thus  produce  a 
vertical  oscillation.  Let  a  force  F  elongate  the  bar  from 
B  tob,  and  let  Bb  =  \,  then,  according  to  the  equation  of 
Article  131,  we  have 


;        -        -        •     (1) 

in  which  E,  JT,  and  I  are  constants,  hence,  the  elongation,  Xj, 
varies  directly  as  the  pulling  force,  F.  The  force  neces- 
sary to  produce  an  elongation  equal  to  unity,  will  be 

F  EK  9. 

_  =  „  =  __;         .  .  .      (2) 


248  KINETICS.  [332.] 

which,  gubstituted  in  equation  (1  )  of  Article  325,  making 
*  =  *i»  gives 


.         .(3) 
or,  substituting  the  value  of  F,  gives 


~EK 


in  which,  if  m  =  P-r-ff,  we  have 


Let  X  be  the  elongation  due  to  P,  and  we  have 

..  (5) 

The  value  of  1;,  equation  (2),  substituted  in  equation  (5)  of 
Article  326,  gives 


I  ml  I  PI 

=  fpVTw?kti»*lV:n 


gEK 

.       .    (6) 

hence,  the  time  of  an  oscillation  is  independent  of  the 
amount  of  elongation  produced  by  the  disturbing  force  F. 

EXAMPLES. 

1.  What  is  the  length  of  a  pendulum  which  will  vibrate 

twice  in  a  second  ? 

2.  What  is  the  length  of  a  pendulum  which  will  vibrate 

once  in  two  seconds  ? 

3.  A  pendulum  whose  length  is   39.1   inches,   vibrates 


[332.J  EXAMPLES.  24S 

86,420  times  in  a  day  ;  how  much  must  it  be  length- 
ened to  vibrate  once  each  second  ? 

4.  A  second's  pendulum  carried  to  the  top  of  a  mountain 

lost  45.5  seconds  in  a  day  ;  required  the  height  of 
the  mountain. 

5.  A  pebble  is  suspended  by  a  fine  thread  two  feet  long ; 

required  the  time  of  making  5  oscillations. 

6.  If  the  radius  of  the  earth  be  20,923,161  feet,  and  g  at 

the  surface  is  32.0902  feet,  what  would  be  the  velo 
city  of  a  body  as  it  passes  the  centre  of  the  earth  if 
it  could  pass  freely  through  it  ? 

7.  In  the  preceding  example,  what  would  be  the  time  of 

passing  from  surface  to  surface  ? 

8.  A  prismatic  bar,  whose  cross-section  is  ^  of  a  square 

inch,  length  5  feet,  coefficient  of  elasticity  28,000,000 
Ibs.,  has  two  weights  suspended  at  its  lower  ends, 
one  of  1,000  Ibs.  and  the  other  of  3,000  Ibs.  The 
latter  weight  suddenly  drops  off ;  required  the  max- 
imum velocity  imparted  to  the  remaining  weight  by 
the  elastic  action  of  the  bar. 

9.  Required  the  time  of  one  vibration  in  the  preceding 

example. 

[In  these  problems  the  mass  of  the  bar  is  neglected.] 


CHAPTER  XVIII. 

GE1TERAL   PROPERTIES    OF   FLUIDS. 

333.  A  fluid  is  a  substance  in  which  its  particles  are 
free  to  move  among  themselves  ;  as  air,  water,  alcohol,  etc. 

A  perfect  fluid  is  a  substance  in  which  the  particles  are 
perfectly  free  to  move  among  themselves,  there  being  no 
friction  nor  cohesion  between  them,  and  in  which  the  least 
force  will  move  any  particle  in  reference  to  surrounding 
particles.  No  snch  substance  is  known  to  exist.  Even  in 
the  most  volatile  gas  its  particles  are  supposed  to  offer 
some  resistance  between  themselves.  But  the  hypothesis 
of  perfect  fluidity  leads  to  results  which  are  useful  in  de- 
termining certain  formulas  applicable  to  imperfect  fluids. 

An  imperfect  or  viscous  fluid  is  one  in  which  there  is  a 
resistance  between  its  particles.  There  are  all  grades  of 
viscosity,  from  that  of  the  most  volatile  gas  to  that  of  solid 
bodies.  Mons.  Tresca,  a  French  physicist,  proved  that 
even  certain  solids,  as  steel  and  iron,  were  somewhat  vis- 
cous. If  steel  be  subjected  to  an  immense  pressure  by  a 
blunt  tool,  the  metal,  in  the  immediate  vicinity  of  the 
place  pressed,  appears  to  flow  like  thick  tar,  or  molasses, 
when  either  is  pressed  at  a  point  on  the  surface.  Several 
armor  plates,  ten  or.  twelve  inches  thick,  which  had  been 
struck  by  cannon  balls,  were  at  the  Centennial  Exhibition, 
and  were  excellent  examples  of  the  viscosity  of  metals. 

There  are  two  classes  of  fluids:  liquids  and  gaseous 
bodies  ;  the  latter  of  which  includes  permanent  gases  and 
vapors,  and  are  called  aeriform  bodies. 


L334-336.]  DEFINITIONS  251 

334.  A  liquid  is  a  fluid  in  which  there  is  a  slight  cohe- 
sion between  its  particles.     Water  is  taken  as  a  type  of 
liquids.     The  Greek  word  for  water  is  "vScop,  hence    the 
science  of  the  statical  equilibrium  of  fluids  is  called  Hydro- 
statics, and  of  their  motion,  Hydrodynamics. 

A  perfect  liquid  is  a  perfect  fluid  in  which  there  is  no 
cohesion  nor  repulsion  between  its  particles.  The  hypo- 
thesis of  perfect  fluidity  is  assumed  unless  otherwise  stated. 

335.  Agriform  "bodies  are  those  in  which  the  particles 
mutually  repel  each  other. 

If  a  vessel,  made  of  elastic  material,  or  provided  with  a 
piston,  be  filled  with  a  gas  and  then  enlarged,  the  gas  will 
constantly  fill  the  vessel.  There  is  no  known  limit  to  the 
expansion  of  a  gas.  The  space  which  it  occupies  depends 
npon  the  pressure  to  which  it  is  subjected. 

336.  Forces  in  the  Three  States  of  Matter. — The 
particles  which  constitute  a  body  act  upon  each  other  by 
forces  of  attraction  and  repulsion.     Supposing  that  both 
forces  exist  at  the  same  time,  the  three  states  of  matter — 
solid,  liquid,  and  gaseous — may  be  defined  by  the  relations 
which  these  forces  bear  to  each  other.     Thus,  a  solid  is  a 
body  in  which  the  force  of  attraction  greatly  exceeds  that 
of  repulsion ;  a  liquid,  one  in  which  the  force  of  attrac- 
tion equals  that  of  repulsion  ;  and  a  gaseous  body,  one  in 
which  the  repulsion  constantly  exceeds  the  attraction. 

Many  solids  may  be  reduced  to  liquids  by  means  of  heat ; 
the  amount  of  heat  depending  upon  the  degree  of  attrac- 
tion existing  between  the  particles.  Thus,  ice,  lead,  zinc, 
iron,  etc.,  are  examples ;  and  in  some  cases  a  substance 
may  be  made  to  assume  the  three  states,  of  which  carbonic 
acid  is  a  well-known  type.  The  repulsive  f crce  may  be 
considered  as  the  effect  of  heat. 


252  FLUIDS.  [337-340.] 

337.  Law  of  Equal  Pressures. — The  pressure  at  any 
point  of  a  perfect  fluid  at  rest  is  equal  in  all  directions. 

If  it  were  not  equal  in  every  direction  the  particle  at 
that  point  would  move  in  the  direction  of  the  resultant  of 
the  forces. 

338.  Normal  Pressure. —  The  pressure  of  a  perfect 
fluid  at  rest  upon  the  surface  of  a  vessel  which  contains 
it  is  normal  to  the  surface. 

For,  if  it  were  not,  it  could  be  resolved  into  two  com- 
ponents, one  of  which  would  be  tangential  to  the  surface 
and  the  other  normal  to  it,  and  the  former  would  produce 
motion. 

339.  Equal  Transmission  of  Pressures. —  If  a   vessel 
contain  a  perfect  fluid  at  rest,  and  the  fluid  be  destitute 
of  weight,  the  pressure  will  be  the  same  at  all  points  within 
the  vessel. 

For  the  pressure  against  the  fluid  arises  from  the  reac- 
tion of  the  sides  of  the  vessel  which  contains  the  fluid,  and  if 
there  was  a  greater  pressure  at  any  point  above  than  at  any 
other  point,  motion  would  result.  Gases  are  so  light,  that, 
for  small  quantities  under  pressure,  their  weight  may  gen- 
erally be  neglected.  Liquids  would  be  without  weight  if 
subjected  to  the  conditions  given  in  Article  77. 

340.  The  pressure  upon  the  lower  part  of  a  vessel  which 
contains  a  heavy  fluid  is  greater  than  that  at  the  upper 
part. 

A  heavy  fluid  is  one  that  has  weight.  The  force  of 
gravity  acts  downward  on  each  particle,  producing  a  down- 
ward pressure.  This  will  be  equally  transmitted  to  every 
point  below  it,  but  not  above  it ;  hence,  the  pressure  below 
any  particular  particle  will  exceed  the  pressure  above  it. 
The  downward  pressure  follows  the  same  general  law  as 


[341,  342. J 


VERTICAL    PRESSURES. 


253 


that  of  a  pile  of  blocks ;  the  pressure  increases  from  the 
top  to  the  bottom. 

341.  In  a  heavy,  perfect  fluid,  every  pressure,  except 
that  due  to  the  weight,  will  be  transmitted  to  every  part 
of  the  vessel  without  diminution  of  intensity. 

This  may  be  shown  experimentally  by  means  of  a  closed 
vessel  provided  with  pistons,  as  in  Fig.  156.  Let  the  ves- 
sel be  filled  with  a  fluid  and  the  pressure  upon  the  pistons 
be  noted.  Then,  if  any  piston  be  pressed  inward  it  will 
be  found  that,  to  prevent  the  other  pistons  from  moving 
outward,  the  same  pressure  per  square  inch  must  be 
applied  to  them  as  to  the  first  piston.  In  other  words,  the 
pressure  per  square  inch  will  be  the  same  on  all  the  pis- 
tons after  deducting  that  due  to  the  weight  of  the  fluid. 


342.  Vertical  Pressures. —  The  difference  between  the 
pressures  on  the  top  and  'bottom  of  a  vertical  prismatic, 
vessel,  filled  with  a  heavy,  perfect  fluid,  equals  the  weight 
of  the  fluid. 

Let  the  narrow  strip  ab,  Fig.  157,  represent  a  vertical 
prism.  If  there  is  a  downward  pressure  on  the  top  at  % 
it  will,  by  the  principles  of  equal  transmission,  be  trans- 
mitted without  diminution  of  intensity  to  b.  The  weight 


254  FLUIJDS.  F343.J 

of  the  particles  will  also  press  downward,  and  since  the 
fluid  is  supposed  to  be  perfect,  the  entire  weight  will  be  sup- 
ported by  the  base  b.  Hence,  the  pressure  at  b  will  equal 
the  downward  pressure  at  a  plus  the  weight  of  the  fluid. 

As  a  result  of  this  proposition,  it  follows  that,  if  there  is 
no  pressure  at  a,  the  pressure  at  b  will  equal  the  weight  of 
the  fluid  in  the  prism.  If  the  vessel  is  filled  with  a  gas  it 
must  be  closed,  and  there  will  necessarily  be  a  pressure  at 
a,  but  in  the  case  of  a  liquid  there  is  not  necessarily  any 
pressure  at  that  point. 

343.  In  Fig.  157,  the  vertical  pressure  upon  the  base  at 
c  is  of  the  same  intensity  as  that  at  b  ;  for  the  pressure  at 
b  will  be  transmitted  horizontally  to  c,  but  at  c  it  will  act 
vertically.  The  pressure  at y  will  be  less  than  that  at  c  by 
the  weight  of  the  fluid  in  the  prism  fo  ;  hence,  the  pres- 
sure on  a  portion  of  the  surface  aty  will  equal  the  weight 
of  a  prism  of  the  fluid  having  for  its  base  the  area  at/J 
and  for  its  height  the  distance  of  the  point  f  below  the 
top  of  the  fluid,  plus  the  downward  pressure  upon  the 
same  area  at  the  top  of  1/he  vessel. 

The  pressure  upon  the  base  of  a  vessel  containing  a 
heavy,  perfect  fluid  is  independent  of  the  form  of  the  ves- 
sel, and  equals  the  weight  of  a  prism  of  the  fluid  having 
for  its  base  the  base  of  the  vessel,  and  for  its  altitude  the 
altitude  of  the  vessel,  plus  a  pressure  upon  each  unit  of 
the  base  equal  to  the  pressure  per  unit  upon  the  upper 
base. 

Let  S  =  the  area  of  the  base  of  the  vessel ; 
a  =  the  altitude  of  the  vessel ; 
B  =  the  weight  per  unit  of  volume  of  the  fluid  ; 
p  =  the  pressure  per  unit  on  the  top  surface  of  the 
fluid ;  and 


[344,345.]  RESOLVED    PRESSURES. 

P  =  the  total  pressure  upon  the  base  ; 
then 


255 


344.  Resultant  pressure  against  the  inside  of  a  ves- 
sel containing  a  heavy  fluid. 

In  Fig.  157,  if  the  points  d  and  e  are  in  the  same  hori- 
zontal, and  directly  opposite  to  each  other,  the  pressure 
upon  them  will  be  equal  and  opposite,  and  similarly  for 
all  other  pairs  of  points  on  the  inside  of  the  vessel ;  hence 
the  resultant  pressure  upon  the  whole  in- 
terior surface  is  zero. 

If  a  hole  be  cut  in  one  side  of  the  vessel, 
then,  as  the  fluid  is  being  discharged,  there 
will  be  no  pressure  on  that  part  of  the  ves- 
sel, and  the  pressure  on  the  part  directly 
opposite  will  tend  to  move  the  vessel  in  the 
direction  of  the  pressure.  If  the  vessel  be 
suspended  by  a  small  cord  the  effect  of  this 
pressure  may  be  observed. 

It  is  on  this  principle  that  sky-rockets  are 
sent  into  the  air.  The  pressure  due  to  the 
burning  of  the  powder  acts  upward  against  the  rocket,  and 
downward  against  the  air,  and  the  former  pressure  raises 
the  rocket. 

345.  Resolved  Pressures. — Let  the   normal   pressure 
on  ABCD  be  uniform  and  equal 

to  P  per  unit,  and  let  6  be  the  in- 
clination of  the  surface  to  the 
vertical ;  then  will  the  horizontal 
component  of  the  pressure  per 
unit  be 

P  =  P  cos  9 


FJQ.  158. 


Fie.  168L 


256 


PRESSURE    OF 


[34G,  :! 


Let  S=  area  ABCD,  and  Sl  =  the  projection  of  the 
area  on  a  vertical  plane  passing  through  AB ;  then 

$  =  #0080, 
and  the  normal  pressure  upon  the  surface  /6\  will  be 


hence,  the  horizontal  component  of  the  pressure  against 
an  oblique  surface  equals  the  horizontal  pressure  against 
the  vertical  projection  of  the  same  surface. 

346.  Resultant  Pressure  on  a  solid  body  immersed 
in  a  heavy  fluid. — The  resultant  of  the  horizontal  pres- 
sures will  be  zero ;  for  the  pressure  on  one  side  of  the 
body  projected  in  a  vertical  plane  (Article  345)  will  equal 
that  on  the  other  side  projected  on  the  same  plane. 

If  the  body  be  divided  into  small  vertical  prisms  be,  the 
vertical  component  of  the  pressure  at  c  will  exceed  that  at 
b  by  an  amount  equal  to  the  weight  of  a 
prism  of  the  fluid  whose  volume  equals 
that  of  the  prism  be  (Article  342).  Hence, 
the  total  upward  pressure  upon  the  body 
equals  the  weight  of  a  quantity  of  the 
liquid  of  the  same  volume  as  that  of  the 
solid.  If  the  weight  of  the  body  be  less 
than  this  pressure,  the  body  will  ascend,  as  is  the  case  with 
a  balloon  rising  in  the  air,  and  light,  wood  rising  in  water ; 
but  if  the  body  be  heavier  than  an  equal  volume  of  the 
fluid  it  will  fall  in  it,  as  in  the  case  of  bodies  falling  in  the 
air,  or  a  stone  sinking  in  water. 

347.  The  point  of  application  of  the  resultant  pressure 
will  be  at  the  centre  of  gravity  of  the  solid,  considered  as 
a  homogeneous  fluid.     If  the  solid  be  not  homogeneous, 
its  centre  of  gravity  will  not  coincide  with  the  centre  of 
pressure,  and  if  such  a  body  be  placed  in  the  fluid  so  that 


FIG.  160. 


[848.] 


FLUIDS. 


257 


its  centre  of  gravity  and  the  centre  of  pressure  are  not  in 
the  same  vertical,  the  body  will  rotate  more  or  less  as  it 
rises  or  falls  in  the  fluid  ;  for  the  two  forces  constitute  a 
couple  (Article  183),  or  a  couple  and  a  single  force  (Arti- 
cles 188,  192).  If  such  a  body  is  not  spherical,  the  com- 
bined motions  of  rotation  and  translation  will  generally 
cause  the  body  to  describe  a  curved  path. 

348.  Equilibrium  of  Fluids  of  Different  Densities.  — 
If  two  fluids  which  do  not  mix  be  placed  in  two  open  ves- 
sels which  communicate  with  each  other,  the  heights  to 
which  they  will  stand  above  their  common  base  will  be  in- 
versely proportional  to  their  densities. 

Let  CabB  be  the  vessel,  A  the  common 
base,  C  the  surface  of  one  fluid  and  B  that 
of  the  other.  The  portion  AD  will  be  in 
equilibrium,  hence  the  pressure  of  BD  will 
equal  that  of  CA. 

Let  8  =  the  density  of  the  liquid  in  BD, 
8^^  =  the  density  of  the  liquid  in  CA, 
h  =  the  height  £D,  and 


=  the  height  CA. 


The  pressure  upon  a  unit  of  area  of 
the  section  at  A,  due  to  the  fluid  in  BD,  will 
be  (Articles  342,  343) 

and  the  pressure  due  to  that  in  AC  will  be 
but,  there  being  equilibrium,  we  have 


FIQ    161. 


25b'  PRESSURE    OF  [348.1 

This  proposition  does  not  apply  to  the  case  hi  which  one 
of  the  fluids  is  lighter  than  air,  for  the  vessel  containing 
the  lighter  fluid  must  be  closed  at  the  top. 

EXAMPLES. 

1.  A  rectangular  closed  box,  whose  depth  is  1  ft.,  breadth 

2  ft.,  and  length  3  ft.,  is  filled  with  a  fluid.  A  cylin- 
drical piston,  whose  diameter  is  one  inch,  is  fitted 
into  the  top  of  the  box ;  required  the  pressure  on 
the  bottom  and  sides  of  the  box  which  would  result 
from  a  pressure  of  20  Ibs.  on  the  piston. 

2.  If  a  vessel  whose  base  is  6  square  inches  and  height  8 

inches,  is  filled  with  water,  what  will  be  the  pressure 
upon  the  base,  calling  the  weight  of  water  62^  Ibs. 
per  cubic  foot  ? 

3.  The  lower  part  of  a  vessel  is  a  cylinder  whose  diameter 

is  8  inches  and  height  6  inches;  the  upper  partis 
also  a  cylinder  whose  diameter  is  6  inches,  and  height 
4  inches,  the  vessel  is  filled  with  water  and  subjected 
to  a  pressure  of  100  Ibs.  on  its  upper  surface ;  required 
the  pressure  on  the  base. 

4.  A  cubic  foot  of  wood  that  weighs  35  Ibs.  is  placed  in  a 

vessel  of  water  which  weighs  63  Ibs.  per  cubic  foot, 
and  the  body  is  prevented  from  rising  by  a  string 
fastened  to  the  bottom  of  the  vessel ;  required  the 
tension  of  the  string. 

5.  In  the  preceding  example,  if  the  body  is  free  to  rise, 

and  there  were  no  resistance  to  motion  from  the  fluid, 
what  would  be  its  velocity  when  it  has  risen  50  feet? 

[The  acceleration  may  be  found  by  Article  86  and  the  velocity 
by  equation  (3),  Article  24.  The  result,  however,  is  of  no  prac- 
tical value,  for  the  resistance  of  the  liquid  will  be  considerable, 
varying  nearly  as  the  square  of  the  velocity.] 


[348.]  FLUIDS.  259 

6.  If  a  cubic  block  of  stone,  whose  edges  are  each  1-|  ft 

and  weight  180  Ibs.  per  cubic  foot,  is  suspended  ir 
water  and  held  by  a  cord,  what  will  be  the  tension 
of  the  cord,  the  water  weighing  62£  Ibs.  per  cubic 
foot  ? 

7.  One  half  of  a  prismatic  bar  is  composed  of  wood  and 

the  other  half  of  iron  ;  the  iron  being  8  times  as 
heavy  as  the  wood,  and  the  whole  immersed  in  a 
liquid  whose  weight  is  65  Ibs.  per  cubic  foot,  at  what 
distance  from  the  middle  must  a  cord  be  attached  so 
that  the  bar  may  rest  in  a  horizontal  position  ? 

8.  If  a  prismatic  tube  is  bent  as  in  Fig.  161,  and  filled 

with  mercury  to  a  height  DA,  how  many  inches  of 
water  must  be  placed  in  the  tube  AC  to  depress  the 
mercury  three  inches,  the  weight  of  mercury  being 
times  that  of  water. 


EXERCISES. 

1.  If  a  vessel  filled  with  water  were  placed  at  rest  in  a  hollow  space  at 

the  eentre  of  the  earth  (see  Article  77),  and  the  vessel  should  sud- 
denly vanish,  would  the  liquid  disperse  ?  would  it  remain  in  the 
same  form  as  that  of  the  vessel  before  it  vanished  ? 

2.  In  the  preceding  exercise,  if  the  vessel  were  filled  with  a  gas,  what 

would  become  of  it  if  the  vessel  should  vanish  ? 

3.  If  a  pail  were  filled  with  tar  would  the  pressure  on  the  bottom  of  the 

pail  equal  the  weight  of  the  tar  ? 

4.  How  much  less  will  a  heavy  body  weigh  in  air  than  it  will  :'n  a 

vacuum  ? 

5.  What  must  be  the  weight  of  a  body  BO  that  it  will  neither  rise  uoi 

fall  in  air? 


CHAPTER  XIX. 

SPECIFIC   GRAVITY. 

349.  Definitions. — The  specific  gravity  of  a  body  is  the 
ratio  of  the  weight  of  the  body  to  the  weight  of  an  equal 
volume  of  some  other  body  taken  as  a  standard.  The 
specific  gravity  of  the  standard  is  taken  as  unity. 

Distilled  water  is  generally  taken  as  the  standard  of 
comparison  for  solids  and  liquids,  and  atmospheric  air  for 
aeriform  bodies,  but  both  of  these  substances  change  their 
volume  for  every  change  of  temperature  and  of  pressure. 
It  is  necessary  to  fix  a  standard  temperature  and  pressure.* 
Some  writers  have  assumed  60°  F.  for  the  standard  tem- 
perature for  water,  while  others  have  taken  it  at  38.75°  F., 
assuming  that  water  at  that  temperature  has  its  maxi- 
mum density.  We  will  assume  the  latter  as  the  tempera- 
ture for  the  standard,  although  the  exact  temperature  cor- 
responding to  the  maximum  density  of  water  is  not 
positively  known,  it  being  fixed  by  some  at  38.85°  F.,  and 
by  others  at  39.101°  F.  The  pressure  of  the  air  is  deter- 
mined by  means  of  a  barometer,  and,  at  the  level  of  the 
soa,  equals  that  of  a  column  of  mercury  about  29.92  inches 
high.  "When  the  pressure  and  temperature  are  known,  the 
specific  gravity  may  be  reduced  to  the  standard. 

The  specific  gravity  of  air  at  32°  F.,  with  the  barometer 
at  30  inches,  is  about  y^,  water  being  unity.  The  exact 
relation  being  established,  all  substances,  including  gases 
and  vapors,  may  be  compared  directly  with  water  as  a 
standard. 


[350,351.]  SPECIFIC    GRAVITY.  261 

[The  term  density,  as  used  in  Mechanics,  is  not  identical  with 
that  of  specific  gravity,  although  the  ratio  of  the  specific  gravi- 
ties of  two  bodies  is  the  same  as  that  of  their  densities.  The 
specific  gravity  of  a  oubic  foot  of  distilled  water  is  unity,  but  its 
density  is  62^  Ibs.  -r-  33  fe,  see  Article  85.] 

350.  To  find  the  Specific  Gravity  of  a  Body  more 
dense  than  that  of  Water. — Weigh  the  body  in  a  vacuum 
and  then   in  the  standard    water ;  let 

w  =  the  former  weight  and  wl  the  lat- 
ter ;  then,  according  to  Article  346,  the 
weight  of  a  quantity  of  water  equal  in 
volume  to  that  of  the  body  will  be 

W  —  Wi,  FIG.  162. 

and,  according  to  the  preceding  article,  the  specific  gravity 
will  be 

w        _  absolute  weight 
w  —  w^        loss  of  weight  ' 

351.  To  find  the  Specific  Gravity  of  a  Body  less 
dense  than  Water. — Attach  it  to  a  body  12,  which  will 
cause  it  to  sink  in  the  water,  and  let 

w  =  the  absolute  weight  of  the  given  body, 
w±  =  the  absolute  weight  of  the  body  \B, 
w2  =  the  absolute  weight  of  both  bodies, 
Wi  =  the  weight  of  B  in'  water,  and 
w2'  =  the  weight  of  the  combined  bodies  in  water. 

Then 

Wi  —  Wi  =  loss  of  weight  of  J$, 

w2  —  w2'  =  loss  of  both  bodies,  and 

(wz  —  w2')  —  (wi  —  Wi)  =  loss  of  weight  due  to  the 
given  body,  which  equals  the  weight  of  a  mass  of  water  of 


262  FLUIDS.  [352,353.] 

the  same  volume  as  that  of  the  given  body.     Substitute  iu 
this  expression 

w%  =  w  -f-  wt, 
and  it  becomes 

W  +  Wjf—  W% 

w 


352.   To  find  the  Specific  Gravity   of  a  Liquid.  — 

Weigh  a  body  in  a  vacuum,  the  water,  and  in  the  required 
liquid. 

Let  w  =  the  weight  of  the  body  in  a  vacuum, 

w^  =  the  weight  of  the  same  body  in  water,  and 
w2  =  the  weight  of  the  same  body  in  the  liquid  ; 
then 

w  —  Wi  =  the  weight  of  an  equal  volume  of  water, 

and 
w  —  w2  =  the  weight  of  an  equal  volume  of  the 

liquid  ; 
but  the  volumes  being  equal,  we  have,  from  the  definition, 

w  —  w2 


w  — 


If  an  empty  bottle,  whose  weight  is  w,  weighs,  when  filled 
with  water,  w1}  and,  when  filled  with  the  liquid,  w2,  we 
have 


—  w 


which  is  the  same  as  the  preceding  formula. 

353.    To  find  the  absolute  Weight  of  a 

Weigh  the  body  in  air  and  in  water. 


[354.]  SPECIFIC    GRAVITY.  263 

Let  Wi  =  the  weight  in  air, 

wz  =  the  weight  in  water,  and 
w  =  the  required  weight  in  a  vacuum ; 
then 

w  —  wt  =  the  weight  of  a  mass  of  air   equal  in 

volume  to  that  of  the  body,  and 
w  —  wz  =  the  weight  of  an  equal  volume  of  water ; 
then,  if  s  be  the  specific  gravity  of  air  compared  with 
water  as  a  standard,  we  have 


8  = 


w  — 


.'.W  = 


1-8 

But  as  *  is  very  small,  less  than  -^,  the  value  of  w  foi 
most  solids  will  be  very  nearly  equal  to  wl ;  hence,  for 
most  practical  purposes,  the  weight  in  air  may  be  used  in- 
stead of  the  absolute  weight. 

354.  Specific  Gravity  of  a  Soluble  Body. — Find  its 
specific  gravity  in  respect  to  some  liquid  in  which  it  is  not 
soluble,  then  find  the  specific  gravity  of  the  liquid  in 
reference  to  water.  Let 

8  =  the  specific  gravity  of  the  liquid  in  reference 

to  water ; 

s{  =  the  specific  gravity  of  the  substance  in  refer- 
ence to  the  liquid  ;  and 

s%  =  the  specific  gravity  of  the  substance  in  refer- 
ence to  water ; 
then 

*2  —  si  • s  5 

for  if  the  body  is  ^  times  as  heavy  as  the  liquid,  and  the 
liquid  s  times  as  heavy  as  water,  then  will  the  substance 
be  «!  times  s  times  as  heavy  as  water. 


264  FLUIDS.  [  355-357.  j 

355.  Specific   Gravity  of  the  Air. — Weigh  a  large 
globe  which  is  filled  with  air.     Exhaust  the  air  as  com- 
pletely as  possible,  the  degree  of  exhaustion  being  deter- 
mined by  a  barometric  column,  and  weigh  again.     Weigh 
the  same  filled  with  water.     Determine  the  total  weight 
of  the  air  originally  in  the  vessel,  and  divide  it  by  the 
weight  of  the  water. 

This  explanation  is  intended  to  give  only  a  very  crude 
idea  of  how  it  may  be  determined.  In  determining  the 
specific  gravity  accurately  there  are  many  details,  a  de- 
scription of  which  is  not  suited  to  this  work. 

Hydrometers,  or  Areometers. 

» 

356.  Instruments  for  determining  the  specific  gravity  of 

fluids,  are  called  Hydrometers,  or  Areometers.  They  are 
of  two  kinds,  one  in  which  the  weight  is  constant  and  the 
other  in  which  the  volume  is  constant. 

357.  Areometer  of  Constant  Weight. — When   the 
same  instrument  is  placed  in  liquids  of  different  densities  it 

will  sink  to  different  depths,  because  the  weight 
of  the  volumes  displaced  must  constantly  equal 
the  weight  of  the  instrument.  In  Fig.  163,  let 
CD  be  a  tube  of  uniform  size,  B  a  hollow  ball, 
and  A  a  small  vessel  containing  mercury,  so  as  to 
make  the  instrument  stand  upright  in  the  fluid. 
To  graduate  the  stem,  place  sufficient  mer- 
cury in  the  vessel  A  so  as  to  make  the  instru- 
ment float  to  some  definite  point  and  mark  it 
1.0.  Then  float  the  instrument  in  a  liquid  whose 

Fio.  163. 

specific  gravity  is  known  to  be  1.1  and  mark  the 
point  to  which  it  sinks  1.1.  Divide  the  intermediate  space 
into  10  equal  divisions,  and  continue  the  divisions  both 


o.i 


[357.J  SPECIFIC    GRAVITY.  265 

above  and  below  as  far  as  desired.  This  method  of  equal 
divisions  is  not  exactly  correct,  as  may  be  seen  from  the 
following  formula.  Let 

V  =  the  volume  of  the  part  aminersed  in  water ; 
v  =  the  volume  included  between  two  consecutive 

divisions  of  the  stem ; 
Di  =  the  density  of  water ; 
D  =  the  density  of  the  liquid ; 
x  =  the  number  of  divisions  between  the  point 
marked  for  water,  and  the  surface  when  it 
floats  in  the  liquid,  and 
8  =  the  specific  gravity  of  the  liquid. 
Since  the  weights  of  the  liquids  displaced  are  constant,  we 
have 


T 
_D_          V 


Let  the  instrument  be  immersed  in  a  liquid  of  known 
specific  gravity,  say  1.1,  and  call  it  *15  and  let  x  =  10  for 
the  space  observed,  and  call  its  value  x^ ;  then  we  find 
from  the  preceding  formula 

V  * 

—  =  — ^r  «io- 

V         8l—\ 

Substituting  this  value  in  the  preceding  equation  gives 


and  letting  x  =  1,  2,  3,  etc.,  the  values  of  the  specific  grav 


266  FLUIDS.  [358., 

ity  corresponding  to  the  successive  spaces  may  be  com- 
puted and  marked  on  the  scale  ;  or,  if  desired,  the  specific 
gravities  may  be  assumed  and  the  spaces,  x,  computed. 

358.  Nicholson's  Hydrometer. — This  is  an  areometer 
of  constant  volume.  It  consists  of  a  hollow  brass  cylinder 
-4,  having  a  small  basket,  B,  at  the  lower  end 
and  carrying  a  small  scale  pan,  E,  at  the  upper 
end.  At  C  is  a  small  vessel  of  mercury  to 
make  the  instrument  float  upright.  Sufficient 
mercury  is  placed  in  the  vessel  so  that  with  500 
grains  in  the  pan,  E,  the  instrument  will  sink 
to  a  given  notch,  D.  This  instrument  may  be 
used  for  determining  the  specific  gravity  of  solids 
or  liquids.  If  the  solid  is  lighter  than  water, 
the  vessel  at  B  is  inverted  so  as  to  force  the  body 
down  into  the  liquid. 

To  find  the  specific  gravity  of  a  solid,  place  a  small 
quantity  of  it  in  the  pan  E  and  add  weights  sufficient  to 
sink  the  instrument  to  D.  Then  place  the  substance  in 
the  basket  B^  and  the  additional  weights  necessary  to  sink 
the  instrument  to  the  same  mark  will  be  the  weight  of  an 
equal  volume  of  water.  Let 

w  =  the  weight  necessary  to  sink  the  instrument 
to  D  in  pure  water ; 

Wi  =  the  weight  which  must  be  added  to  the  sub- 
stance to  sink  the  instrument  to  the  same 
point,  when  the  substance  is  in  the  pan  E\ 

w%  =  the  weights  in  the  pan  E,  when  the  substance 
is  in  the  basket  B,  necessary  to  sink  it  to 
the  same  point,  and 
8  =  the  specific  gravity  of  the  substance. 


[859.]  SPECIFIC    GRAVITY.  267 

Then 

w  —  «?j  =  the  weight  of  the  substance  in  air, 

w2  —Wi  =  the  loss  of  weight  of  the  substance  in  water ; 

w  —  Wi 


.To  find  the  specific  gravity  of  a  liquid.  Sink  the  in- 
strument to  the  same  depth,  D}  in  water  and  in  the  liquid, 
and  let 

W=  the  weight  of  the  instrument, 
w  =  the  weight  in  the  pan  E  when  in  water,  and 
Wi  =  the  weight  in  the  pan  when  in  the  liquid  ; 
then 


_ 
8~W+w' 


Problems. 

359.  Mechanical  Combinations. — To  find  the  weights 
of  the  constituents  in  a  mechanical  composition  when  the 
specific  gravities  of  the  compound  and  the  constituents 
are  known. 

This  is  a  general  statement  of  the  noted  problem  solved 
by  Archimedes,  in  which  he  determined  the  respective 
amounts  of  gold  and  silver  in  King  Hiero's  crown. 

Let  w,  wl}  «?2,  be  the  weights  of  the  compound  and  con- 
stituents respectively ; 

s,  81,  S2,  their  respective  specific  gravities;  and 
V,  Vi,  Vfy  their  respective  .volumes. 

In  mechanical  combinations  we  have 

w  =  Wi  +  Wi ;       .         .        .     (1) 

v  =  Vi+  vz.          .        .        .     (2) 


268  FLUIDS.  [360] 

But 


w  —  gDv  ;  wt  =  gD^  ;  w.z  =  gD&t  ;     .    (3) 
which,  combined  with  equation  (2),  gives 

™_^+?^.  (4) 

D~  A    A' 

or  since  their  specific  gravities  are  as  their  densities, 

»  =  ^+^;  (5) 

o  o  o  ' 

a  f>i          esg 

which,  combined  with  equation  (1),  gives 


l  -  --  } 

(*»  -  *l)« 


360.  Chemical  Combination.  —  Two  fluids  whose  vol- 
umes are  v  and  v^  and  specific  gravities  s  and  st  respect- 

ively, on  being  mixed,  contract  -  t/ipart  of  the  sum  of  their 

iv  _ 

volumes  ;  required  the  specific  gravity  of  the  mixture. 

Let  s2  =  the  specific  gravity  required,  and 

8  =  the  weight  of  a  unit  of  volume  of  water. 

If  there  were  no  condensation  the  volume  after  mixture 
would  be 


but,  on  account  of  the  condensation,  it  will  bo 


[361,362.]  SPECIFIC    GRAVITY.  269 

The  sum  of  their  weights  before  mixture  will  equal  the 
total  weight  after  mixture,  hence, 


n      vs  + 


2~~  n  —  1*    v  +  vt 

361.  In  the  preceding  problem  the  specific  gravity  of 
the  mixture  being  found,  required  the  amount  of  conden- 
sation. 

Solving  for  —  gives 


n 


362.  To  find  the  Specific  Gravity  of  a  Body  lighter 
than  Water  when  -weighed  in  Air.  —  A  body  B±,  whose 
density  is  less  than  water,  weighs  bt  grains  in  air,  and  B^ 
in  water  weighs  b2  grains,  and  B±  and  Z?2  connected,  weigh 
c  grains  in  water.  The  specific  gravity  of  air  being 
0.0013,  required  the  specific  gravity  of  B±. 

Let  v±  and  v^  be  the  volumes  respectively  of  B±  and  B^ 
*!  and  «2  their  specific  gravities,  and 
B  the  weight  of  a  unit  of  volume  of  water. 
Then 

(a,  —  I)0j8  =  52  . 


_    =  c. 

From  these  we  find 

_bi  +  0.0013(J2  —  c) 

8l   7         .      7 1 • 


270  FLUIDS.  [363.] 

EXAMPLES. 

1.  A  piece  of  wood  weighs  12  Ibs.,  and  when  annexed  to 

22  Ibs.  of  lead,  whose  specific  gravity  is  11,  the  whole 
weighs  8  Ibs.  in  water;  required  the  specific  gravity 
of  the  wood.  Ans.  £. 

2.  Required  the  specific  gravity  of  a  body  which  weighs 

32  grains  in  a  vacuum  and  25  grains  in  water. 

3.  An  areometer  sinks  to  a  certain  depth  in  a  fluid  whose 

specific  gravity  is  0.8,  and  when  loaded  with  60 
grains  it  sinks  to  the  same  depth  in  water  ;  required 
the  weight  of  the  instrument. 

4.  A  cubic  foot  of  water  weighs  62£  Ibs. ;  required  the 

weight  of  a  cubical  block  of  stone  whose  edges  are 
each  5  ft.,  its  specific  gravity  being  2.3. 
6.  If  a  body  sinks  £  of  its  volume  in  distilled  water,  what 
is  its  specific  gravity  ? 

6.  A  body,  whose  weight  is  40  grains,  weighs  35  grains  ic 

water  and  32  in  an  acid ;  required  the  specific  grav- 
ity of  the  acid. 

7.  Two  pieces  of  metal  weigh  respectively  5  and  2  Ibs., 

and  their  specific  gravities  are  7  and  9  ;  required 
the  specific  gravity  of  the  alloy  formed  by  melting 
them  together,  supposing  that  there  is  no  condensa- 
tion. Ans.  7.474. 

8.  A  compound  of  gold  and  silver  weighing  10  Ibs.  has  a 

specific  gravity  of  s  =  14,  that  of  gold  being  ^  =  19.3, 
and  of  silver  «2  =  10.5  ;  required  the  weight  wt  of 
the  gold  and  w2  of  the  silver. 

Ans.  wj,  =  5.483  Ibs.,  w2  =  4.517  Ibs. 
0.  If  73  Ibs.  of  sulphuric  acid,  the  specific  gravity  of  which 
JB  1.8485,  are  mixed  with  27  Ibs.  of  water,  and  the  re- 


[362.]  SPECIFIC    GRAVITY  271 

suiting  dilute  acid  has  a  specific  gravity  of  1.6321, 
what  will  be  the  amount  of  condensation  ? 

[Before  the  formula  in  Article  361  can  be  used,  it  will  be 
necessary  to  express  the  quantities  in  terms  of  volumes.  Let 
1  Ib.  of  water  be  a  unit  of  volume,  then  will  the  volume  of  water 
be  v  =  27,  and  of  the  acid  «,  =  73  -t- 1.8485  =  39.4915  ;  and  in 
the  formula,  s  =  1.] 

Am.  0.0785. 

10.  A  body  B^  weighs  10  grains  in  water,  and  Bv  14 
grains  in  air,  and  B^  and  B%  together  weigh  7  grains 
in  water ;  required  the  specific  gravity  of  B^,  that 
of  air  being  0.0013.  Ana.  0.8237. 


EXERCISES. 

1.  If  a  body  floats  at  a  certain  depth  in  a  liquid  when  the  vessel  which 

contains  it  is  in  the  air,  will  it  sink  to  the  same  depth  when  the 
vessel  is  in  a  vacuum  ? 

2.  Why  will  smoke  sometimes  rise  in  the  air  ?    Why  will  it  fall  at  other . 

times  ? 

3.  Will  the  depth  to  which  a  body  floats  in  a  liquid  be  affected  by 

changes  in  the  density  of  the  air  ? 

4.  If  a  rubber  bag  containing  a  gas  be  made  to  just  sink  in  a  liquid, 

will  a  pressure  on  the  surface  of  the  liquid  condense  the  gas  ?  and 
if  so  will  it  have  a  tendency  to  rise  ? 

5.  Considering  the  compressibility  of  iron  and  of  water,  can  iron  sink  so 

deep  in  water  as  to  float  at  that  depth  ?  or,  in  other  words,  will 
the  water  become  as  dense  as  the  iron  ? 

6.  If  water  were  incompressible,   is  there  any  limit  to  the  depth  to 

which  a  body  heavier  than  water,  and  also  incompressible,  will 
sink  in  the  water  ?     If  the  body  were  compressible,  is  there  a  limit  ? 

7.  If  an  egg  will  sink  in  pure  water,  will  it  float  or  sink  in  brine  ?  What 

must  be  the  condition  of  the  brine  that  the  egg  may  float  between 
the  top  and  bottom  ? 

8.  Will  a  vessel  of  water  which  contains  a  fish  weigh  any  more  than  if 

the  fish  were  removed  ? 


CHAPTER  XX. 

HYDROSTATICS. 

363.  Compressibility  of  Liquids. — The  mechanical 
properties  of  liquids  are   determined  on   the   hypothesis 
that  liquids  are  incompressible.     They  are,  however,  more 
compressible  than  most  solids.     If  a  cubic  inch  of  water 
be  pressed  with  fifteen  pounds  on  each  and  every  side,  the 
volume  will  be  diminished  3^^-^,  hence  one  pound  to  the 
square  inch  will  diminish  the  volume  -gu-oVmr-     -^  the  water 
be  confined  in  a  perfectly  rigid,  prismatic  vessel,  the  com- 
pression would  take  place  entirely  in  the  direction  of  the 
length,  and  would  equal  -3-0  <jV<nF  of  the  length  for  every 
pound  per  unit  of  area  of  the  end  pressure.     Water,  there- 
fore, is  nearly  100  times  as  compressible  as  steel.     See 
Article  130.     All  other  liquids  are  more  or  less  compres- 
sible, yet,  for  most  practical  purposes,  they  may  be  con- 
sidered   as  non-elastic  without  involving  sensible   error. 
Liquids  are  sometimes  defined  as  non-elastic  fluids. 

The  first  experiment,  to  determine  the  compressibility 
of  water,  was  made  by  a  philosopher  at  Florence,  Italy. 
He  filled  a  hollow  globe  made  of  gold  with  water,  and 
then  subjected  it  to  a  great  pressure,  thereby  flattening  it. 
This  diminished  the  volume,  and  it  was  observed  that  the 
water  oozed  out  through  the  pores  of  the  gold ;  from  which 
he  drew  the  erroneous  conclusion  that  the  liquid  was  not 
diminished  in  volume. 

364.  Free  Surface. — The  upper  surface   of   a  liquid 
contained  in  a  vessel  which  receives  no  press ure,  is  called 


[365.J  PROBLEMS.  273 

the  free  surface.  The  upper  surface  of  water  in  the 
atmosphere  is  pressed  downward  by  the  air  with  about 
fifteen  pounds  to  the  square  inch  ;  yet  such  a  surface  is 
often  considered  as  a  free  surface. 

The  free  surface  of  small  bodies  of  a  perfect  liquid  at 
rest  may  be  considered  as  horizontal  ;  for  it  will  be  per- 
pendicular to  the  direction  of  action  of  the  force  of  grav- 
ity, Articles  338  and  204  ;  but  for  large  bodies  of  a  liquid 
it  is  spherical,  partaking  of  the  general  form  of  the  sur- 
face of  the  earth. 

365.  A  Level  Surface  is  one  which  cuts  at  right  angles 
the  resultant  of  the  forces  which  act  upon  its  particles. 
Thus,  in  a  vessel  filled  with  a  heavy  liquid  at  rest,  it  is 
horizontal  ;  in  the  ocean  it  may  be  a  surface  at  any  depth 
and  nearly  concentric  with  the  free  surface  ;  in  the  second 
problem  below  it  is  a  paraboloid  of  revolution,  etc. 

Problems. 

1.  A  vessel  is  filled  with  a  perfect  ',  homogeneous  liquid, 
and  drawn  horizontally  with  a  uniform  acceleration  ;  re- 
quired the  form  of  the  free  surface. 

Let  F  be  the  force  producing  an  acceleration  f,  and  M 
the  mass  of  the  liquid.     Then,  according  to  Article  86, 
we  have,  for  the  horizontal 
force, 


•7 

and  for  the  vertical  force, 


W=ob. 


These  forces  will  be  uniformly  distributed  throughout 
the  mass  ;  hence  the  resultant  of  the  forces  on  the  parti 


274 


HYDROSTATICS. 


cles  will  be  equal  and  parallel  to  each  other,  and  also  nor- 
mal to  the  free  surface,  Article  338 ;  therefore  the  free 
surface  will  be  a  plane.  The  level  surfaces  will  also  be 
planes  parallel  to  the  free  surface. 

Let  <f>  be  the  inclination  of  the  free  surface  to  the  hori 
zontal,  also  =  cob  ;  then 

tang  d>  =  — ^  =  ^. 
W        g 

2.  If  a  cylindrical  vessel  containing  a  perfect,  homoge- 
neous liquid  be  revolved  uniformly  about  a  vertical  axis, 
what  will  be  the  form  of  the  free  surface  f 

The  vessel  may  be  any  solid  of  revolution,  the  axis  of 
revolution  coinciding  with  the  axis  of  rotation.  Any 

element  of  the  liquid  will  be  acted 
upon  by  two  forces ;  one,  the  force 
of  gravity  acting  vertically  down- 
ward and  equal  to  ob,  the  weight, 
w,  of  the  particle  ;  the  other,  the 
centripetal  force,  acting  horizon- 
tally and  radially  inward.  Let  the 
vessel  be  at  rest  and  a  force  equal 
to  the  centrifugal  force  act  upon 
Fia- 166-  the  particles. 

Let  a)  be  the  angular  velocity,  and  r  the  distance  of  any 
particle  from  the  axis  of  rotation;  then,  according  tc 
Article  314,  the  centrifugal  force  will  be 

w      2       x 
oa  =  mrtar  =.  —  rar  =  oc ; 

9 


and  the  vertical  force, 


ob  —w. 


[365.]  PROBLEMS.  275 

The  resultant  of  these  forces  must  be  normal  to  the  free 
surface.  Prolong  the  line  of  the  resultant  co,  until  it 
meets  the  axis  at  D\  then,  from  the  similar  triangles  obc 
and  o  CD,  we  have 

DC  _  ob 
Co  ~  cV 
or 

•    DC  _     w 

r     ~  w 

—  rar 


or 

that  is,  the  subnormal,  D  C,  is  constant.  It  is  shown  in  the 
Calculus  that  the  parabola  is  the  only  curve  which  pos- 
sesses this  property ;  hence  the  surface  is  a  paraboloid  of 
revolution.  All  the  level  surfaces  are  equal  paraboloids. 

It  is  also  shown  in  the  Calculus  that  the  volume  of  a 
paraboloid  of  revolution  is  one-half  that  of  a  circumscribed 
cylinder ;  hence,  if  the  cylinder  be  at  rest,  the  free  sur- 
face will  be  midway  between  the  highest  and  lowest 
points  of  the  paraboloid. 

3.  If  a  perfect,  homogeneous  mass  of  liquid  be  acted 
upon  by  a  force  which  varies  directly  as  the  distance  from 
the  centre  of  the  mass,  what  will  be  the  form  of  the  free 
surface  f 

It  will  be  a  sphere ;  for  the  force  at  the  surface  will 
then  be  equal  and  normal  at  every  point  of  it. 

4.  If,  in  the  preceding  example,  the  mass  rotates  about 
an  axis,  what  will  be  the  form  of  the  free  surface  f 

The  force  of  gravity  will  act  directly  towards  the  cen- 
tre of  the  mass,  and  the  centrifugal  force  wirf  act  outward, 


276  HYDROSTATICS.  [365.] 

perpendicular  to  the  axis  of  rotation,  and  the  restVtant  of 
these  forces  must  be  normal  to  the  surface.  This  prob- 
lem is  approximately  that  of  the  earth,  and  its  solution 
involves  higher  mathematics.  The  form  is,  approximately, 
an  ellipsoid  of  revolution,  and  iis  often  called  an 
spheroid. 


EXAMPLES. 

1.  A  vessel  containing  a  liquid,  whose  weight  (including 

the  liquid)  is  50  Ibs.,  is  drawn  horizontally  by  an 
effective  moving  foi*ce  (Article  87)  of  15  Ibs. ;  re- 
quired the  inclination  of  the  surface  to  the  hori- 
zontal. 

2.  A  rectangular  box  3  feet  long  contains  a  quantity  of 

liquid.  If  the  liquid  is  one  foot  deep,  what  must  be 
the  acceleration  of  the  box  in  a  horizontal  direction 
that  the  free  surface  at  the  forward  end  shall  just 
touch  the  bottom  of  the  vessel  ? 

3.  In  the  preceding  example,  if  the  rear  end  of  the  box 

slopes  outward  at  an  angle  of  45  degrees,  what  must 
be  the  acceleration  of  the  box  so  that  all  the  water 
shall  escape  by  flowing  over  that  end  ? 

4.  In  Fig.  166,  if  the  vessel  is  cylindrical  and  2  ft.  in  dia- 

meter, and  the  free  surface  of  the  liquid  is  3  inches 
from  the  top,  what  must  be  the  number  of  turns  pei 
minute  so  that  the  upper  edge  of  the  surface  shall 
just  reach  the  edge  of  the  vessel  ? 

5.  If  the  vessel  is  rotated  30  turns  per  minute,  what  will 

be  the  equation  of  the  parabola  ? 

An*,  v1  =  2      x. 


[366.] 


LAW    OF    PRESSURE. 


277 


Law  of  Pressure. 

366.  The  pressure  of  a  perfect,  homogeneous  liquid 

varies  directly  as  the  depth  below  the  free  surface. 

Since  such  a  liquid  is  incompressible,  we  may  consider 
a  vertical  prism  of  the  liquid  as  composed  of  blocks  of  equal 
size  and  weight,  placed  one  above  the  other.  The  first 
block  will  press  with  its  entire  weight  upon  the  second 
one,  and  the  first  and  second  upon  the  third,  and  so  on,  and 
since  the  weights  are  equal  to  each  other,  the  pressure 
upon  the  succeeding  blocks  will  vary  as  the  number  of  the 
blocks,  or  as  1,  2,  3,  etc. 

Draw  a  horizontal  line,  1#,  to  represent  the  pressure  at  1, 
then  will  2&,  representing  the  pressure  at  2,  be  twice  as 
long  as  \a ;  3o,  three  times  as  long,  and  so  on.  In  this  case 


,. 


-iff 


Y 


Fio.  168. 


the  pressure  does  not  increase  continuously,  but  by  steps. 
If  now  the  blocks  be  divided  indefinitely,  the  steps  will 
become  indefinitely  small,  and  ultimately  may  be  repre- 
sented by  Fig.  168,  in  which  AE  represents  the  pressure 
at  J.,  and  any  horizontal  line  drawn  from  AB  to  JBE,  the 
pressure  at  that  point.  Thus  far  we  have  considered  the 
vertical  pressure  only,  but  in  a  perfect  liquid  the  pressure 


278  HYDROSTATICS.  [367,  368.  j 

will  be  the  same  in  all  directions,  Article  337  ;  hence  the 
truth  of  the  proposition.     It  follows  from  this  that  : 

367.  The  pressure  against  an  elementary  area  equals 
the  weight  of  a  prism  of  the  liquid  whose  base  is  the  area 
pressed  and  whose  altitude  is  the  vertical  distance  of  the 
area  below  the  free  surface. 

Let  Aa  =  be  the  area, 

h  =  the  distance  below  the  free  surface,  and 
8  =.  the  weight  of  a  unit  of  volume  ; 

then  the  pressure  will  be 

B.  Aa.h. 

368.  To  find  the  pressure  of  a  liquid  against  a  ver- 
tical rectangle  in  which  one  edge  coincides  with  the  free 
surface. 

Let  ABOD  represent  the  rectangle,  in  which  the  side 
B  C  coincides  with  the  surface  of  the  liquid.  Draw  the 
horizontal  line  AE  to  represent  the 
pressure  at  A,  and  draw  BE\  then 
will  the  triangle  ABE  represent 
the  pressure  against  the  line  AB. 
Complete  the  triangular  wedge 
ABCD—FE;  the  volume  of  this 
wedge  will  represent  the  entire  pres- 
sure against  the  rectangle. 

Let  6  =  the  weight  of  a  unit  of  volume  of  the  liquid  ; 
h  -  AB  ;  b  =  AD. 

The  pressure  on  a  unit  of  area  at  A  will  be,  according  to 
Article  366, 


hence,  the  volume  of  the  wedge  will  l/e 


[369,370.] 


LAW  OF   PRESSURE. 


279 


bh.tfh  =  tfbh\ 

If  the  liquid  be  water,  this  expression  becomes 


in  which  b  and  A  are  in  feet. 

We  see  from  this  expression  that  the  entire  pressure 
from  the  free  surface  downward,  varies  as  the  square  of 
the  distance  from  the  surface. 

369.  To  find  the  pressure  of  a  liquid  against  a  vertical 

rectangle  when  the  upper  edge  is  parallel  to  the  free 
surface. 


l=AD-, 

then  will  the  pressure  on  the  rect- 
angle ABCD  equal  the  difference 
of  the  pressures  on  ANMD  and 
CBNM,  or 


-  A,2). 

370.  Pressure  on  any  Surface. — Conceive  the  surface, 
whether  plane  or  curved,  to  be  divided  into  small  areas. 

Let  «u  «2,  ss,  etc.,  be  the  areas  ; 

A!,  A2,  AS,  etc.,  their  respective  distances  below  the 

free  surface ; 
S  =  81  +  s2  +  ss  H-  etc.  —  5*,  be  the  entire  area  of 

the  surface  pressed  by  the  liquid  ; 
X,  the  depth  of  the  mean  pressure ;  and 
8,  the  weight  of  a  unit  of  volume. 

Then,  according  to  Article  367,  the  pressure  upon  the  sur 
face  will  be 

4- 


260 


HYDROSTATICS. 


[370.] 


and  this  also  equals  the  total  area  into  the  mean  pressure, 
hence 

B&c  =  SSsh  ; 


S  ' 

hence,  according  to  Article  216,  x  is  the  depth  of  the 
tre  of  gravity  of  the  surface.  Therefore,  the  total  pressure 
on  any  surface,  S,  equals  the  weight  of  a  prism  of  the 
liquid  whose  base  is  the  area  of  the  surface  pressed,  and 
whose  altitude  is  the  depth  of  the  centre  of  gravity  of  the 
surface  below  the  free  surface. 

Problems. 

1.  Triangular  Surfaces.  —  To  find  the  pressure  against 
a  triangular  surface  whose  base  is  parallel  to  the  free 
surface,  and  whose  apex  is  in  that  surface. 

Let  ABC  be  the  triangle,  b  =  AC,  h  =  BD,  and  g  the 


centre  of  gravity  of  the  triangle  ;  then,  according  to  Arti- 
cle 370,  we  have 


The  pressure  is  also  represented  by  the  volume  of  the 
pyramid  B—ACFE,  or 


as  before. 


[370.  J  PROBLEMS.  281 

2.  Let  the  base  coincide  with  the  free  surface.     Then 
we  have 


This  may  also  be  represented  by  a  triangular  pyramid  whose 
base  is  A£C\  Fig.  172,  and  whose  altitude  is  AJ?=  Sk. 

3.  Cones. — Find  the  normal  pressure  upon  the  concave 
surface  of  a  closed  cone  filled  with  a  liquid  /  (1)  with  the 
axis  vertical  and  apex  uppermost  :  (2)  axis  vertical  and 
cone  inverted  /  (3)  with  the  axis  horizontal  /  (4)  pressure 
on  the  base  in  case  (1)  /  (5)  vertical  pressure  on  the  con- 
cave  surface  in  (1)  /  (6)  weight  of  the  liquid. 

Let  r  =  the  radius  of  the  base,  and 

h  =  the  altitude  of  the  cone. 
Then 

(1)         4WWAVV  +  A8.  (4)         STT^A. 


(2)  ^TrrhV^+h*.  (5) 

(3)  STn^VV  +  A2.  (6) 


Observe  that  the  weight  equals  (4),  the  downward  pres- 
sure, minus  (5)  the  upward  pressure. 

4.  Spheres. — A  sphere  is  submerged  in  a  liquid :  find 
the  normal  pressure  upon  the  external  surface  (1)  when  it 
is  just  submerged ;  (2)  when  submerged  to  any  depth  ; 
(3)  weight  of  a  quantity  of  the  liquid  equal  in  volume  to 
that  of  the  sphere. 

Let  r  =  the  radius  of  the  sphere,  and 

h  =•  the  depth  of  the  centre  of  the  sphere  below  the 

free  surface. 
Then 

(1)    4&n*;        (2)    4&m»h ;        (3) 


282  HYDROSTATICS.  [371.J 

EXAMPLES. 

1.  In  Fig.  170,  if  the  edge  MN  of  the  rectangle  coincides 

with  the  surface  of  the  liquid,  and  AN  is  3  feet, 
how  far  from  the  surface  must  the  line  CJS  be  drawn 
so  that  the  pressure  on  the  two  parts  shall  be  equal  ? 

Ans.  2.121  feet. 

2.  A  rectangle  whose  sides  are  \A  feet  and  2.6  feet  re- 

spectively, is  immersed  in  water  with  the  former 
side  in  the  surface,  and  is  inclined  at  an  angle  of 
56°  35'  to  the  free  surface  ;  required  the  pressures 
on  the  parts  into  which  the  rectangle  is  divided  by 
its  diagonal. 

3.  A  cylinder  whose  base  is  2  feet  in  diameter  and  alti- 

tude 3  feet,  is  filled  with  water;  required  the  pres- 
sure on  the  concave  surface,  the  pressure  on  the 
base,  and  the  weight  of  the  water. 

4.  A  sphere  10  feet  in  diameter  is  filled  with  water ;  re- 

quired the  normal  pressure  on  the  interior  surface, 
and  the  weight  of  the  fluid. 

5.  Find  the  pressure  on  a  rectangular  submerged  flood- 

gate, ABCD,  Fig.  170,  whose  depth,  BN,  is  10  ft, 
height  of  the  gate,  AB,  3  ft.,  and  width,  £C\  2  ft. 

6.  In  the  preceding  example,  find  the  pressure  if  the  top 

of  the  floodgate  is  also  submerged  on  the  opposite 
side  to  a  depth  of  4  feet. 

Centre  of  Pressure. 

371.  The  centre  of  pressure  of  any  surface  immersed 
ill  a  fluid  is  the  point  of  application  of  the  resultant  of  al] 


[373-375.]  CENTRE    OF    PRESSURE.  283 

the  pressures  upon  it.  It  is  therefore,  that  point  in  an 
immersed  surface  to  which,  if  a  force  equal  and  opposite 
to  the  resultant  of  all  the  pressures  upon  it  be  applied, 
this  force  will  keep  the  surface  in  equilibrium. 

372.  Rectangles. — Let  the  surface  be  a  rectangle  in 
which  one  end  coincides  with  the  free  surface  of  the  liquid  ; 
then  will  the  centre  of  pressure  be  at 

two-thirds  the  depth  of  the  rectangle. 
The  total  pressure  may  be  represented 
by  a  wedge  whose  end  is  the  triangle 
ABE.  Hence  the  centre  of  pres- 
sure will  be  at  the  same  depth  as  that 
of  the  centre  of  gravity  of  the  triangle 
ABE.  Let  6y,  on  the  vertical  line 

rIOK  lio. 

BA,  be  on  a  horizontal  line  through 

the  centre  of  gravity,  g,  of  the  triangle  ABE.     Then, 

according  to  Article  222,  BO  will  be  \BA. 

373.  Submerged  Rectangle. — In  Fig.  170,  the  centre 
of  pressure  will  be  at  the  depth  of  the  centre  of  gravity  of 
the  trapezoid  ABGE.     Since  BG  and  AE are  directly 
proportional  to  NB  and  JVA,  we  have,  from  Example  6, 
page  146,  for  the  depth  required, 

0 
NA-^AB 

374.  Triangles. — The  centre  of  pressure  of  the  triangle 
in  Fig.  171,  will  be  opposite  the  centre  of  gravity  of  the 
pyramid  B—A  CFE;  or 


375.  The  centre  of  pressure  against  the  triangle  AB  C, 
when  CB  is  in  the  free  surface,  is  at  the  centre  of  grav- 
ity g,  of  the  wedge  ABC—E.  To  find  this  point,  we 
observe  that  this  wedge  is  what  remains  after  removing 


284 


HYDROSTATICS. 


[376-377.] 


wedge 


the    pyramid   BCGF—E,   Fig.   175,   from    the 
BCGF—AE.     The  centre  of  gravity  of  the  large  wedge 
is  at  one-third  its  altitude,  and  of  the  pyramid  at  one- 


FIG.  175. 

fourth  its  altitude  from  the  base  ;  hence,  according  to  Arti- 
cle 224,  we  find  that  the  centre  of  pressure,  g,  is  at  one-half 
the  altitude  from,  the  base  CB. 

Flotation. 

376.  Plane  of  Flotation. — If  a  body  in  a  liquid  is 
lighter  than  the  liquid,  it  will  float,  and  the  conditions  of 
equilibrium  will  be  determined  according  to  Article  346. 
The  intersection  of  the  plane  of  the  free  surface  with  the 

floating  body  is  called  the  plane 
of  flotation.  The  line  joining 
the  centre  of  gravity  of  the 
solid  and  of  the  displaced 
liquid  is  called  the  axis  of 
flotation. 

377.  Conditions  of  Equili- 
brium of  a  Floating  Body. — 
One  condition  is,  according  to 

Article  346, that  the  weight  of  the  displaced  liquid  shall 
equal  that  of  the  body.  Another  is  that  the  axis  of  flota 
tion  shall  be  vertical. 


Fio.  176. 


[378,379.]  FLOTATION.  285 

378.  Stable  Equilibrium.—  In  Fig.  176,  let  C  be  the 
centre  of  gravity  of  the  displaced  liquid,  and  G  that  of  the 
body  when  the  axis  of  flotation  is  vertical  ;   and  >C'  the 
centre  of  gravity  of  the  displaced  water  when  the  axis  is 
inclined.     Let  the  vertical  through  C'  meet  the  line  G  V 
in  the  point  M.     When  the  body  is  turned  through  an  in- 
definitely  small  angle,  the  point  M  is  called  the  metacen- 
tre.     When  M  is  above  G  the  pressure  of  the  fluid  up- 
wards along  C'M,  and  of  the  body  downwards  along  the 
vertical  G  W,  tend  to  bring  the  body  back  to  the  position 
in  which  the  axis  of  flotation  will  be  vertical.     Hence  the 
equilibrium  is  stable  when  the  metacentre  is  above  the 
centre  of  gravity  of  the  body.     Observing  that  C'  G  is  a 
new  axis  of  flotation,   it  follows  that  the  equilibrium  is 
stable  when  the  axis  of  flotation  turns  in  a  direction  oppo- 
site to  that  of  the  rotation  of  the  body  when  the  position 
of  the  body  is  disturbed. 

379.  Depth  of  Flotation.  —  Let  D  be  the  density  of  the 
body,  V  its  volume,  and  s  its  specific  gravity  ;  and  D^  Fi, 
«1}  the  corresponding  quantities  for  the  displaced  liquid. 
Then,  according  to  Articles  85  and  349,  and  the  first  con- 
dition of  Article  377,  we  have 


If  the  body  floats  in  pure  water  then  *i  =  1  and 


Problems. 

1.  Let  the  body  be  a  right  cone  with  the  axis  vertical 
and  apex  upward  ;  required  the  depth  of  flotation. 


286  HYDROSTATICS.  [379. 

Let  r  =  the  radius  of  the  base, 
h  =  the  altitude,  and 
x  =  the  depth  of  flotation. 
Then 

T 

•j-  (A — a?)  =  the  radius  of  the  plane  of  flotation. 

hence 

V  = 


and  the  equation  of  the  preceding  article  becomes 
<,-frrj-z(?i-xf=%8- 


2.  .4  rectangular  wall  whose  height  is  hfeet,  thickness 
bfeet,  and  weight  of  a  cubic  foot  of  the  masonry  B  pounds, 
resists  the  pressure  of  water  whose  height  behind  the  wall 
is  AX  feet  /  will  the  wall  be  stable  in  reference  to  slipping 
on,  its  base,  or  to  overturning  about  its  outer  edge  f 

The  pressure  of  the  liquid  for  a  unit  of  width  will  be, 
according  to  article  368, 


and  the  centre  of  pressure,  according  to  Article  372,  will 
be  at  -^A!  from  the  base,  hence  the  moment  of  pressure  in 
reference  to  the  outer  edge  of  the  wall  will  be 


The  weight  of  the  wall  will  be 


[379.1  FLOTATION.  287 

and  the  moment  in  reference  to  the  lowest  outer  point 
will  be 


hence,  the  wall  will  be  stable  in  reference  to  rotation,  if 

SJ2A>20«7tl3; 
and  in  reference  to  slipping  on  the  base,  if 


in  which  p,  is  the  coefficient  of  friction  of  the  wall  on  its 
foundation  ;  see  Article  107. 


EXAMPLES. 

1.  In  Fig.  170,  what  is  the  depth  of  the  centre  of  pressure 

below  MJW,  the  end  MJW  coinciding  with  the  free 
surface,  and  AN  being  3  feet?  Ans.  Ifeet. 

2.  In  Fig.  170,  if  ABCD  is  a  floodgate,  BN  being  5  feet, 

and  AB  10  inches,  how  far  below  B  must  a  hori- 
zontal bar  be  placed  so  as  to  balance  the  pressure 
against  the  gate  ? 

3.  If  the  height  of  a  rectangular  wall  be  8  ft.,  weight  of 

the  masonry  180  Ibs.  per  cubic  foot,  what  must  be 
the  thickness  of  the  wall  to  resist  overturning  from 
the  pressure  of  water  when  level  with  the  top  of  the 
wall? 

4.  A  triangular  wall  whose  base  is  4  ft,  height  8  ft, 

weighs  120  Ibs.  per  cubic  foot ;  required  the  height 
of  water  which  it  will  sustain  in  reference  to  over- 
turning, and  leave  a  coefficient  of  stability  of  2. 

[The  degree  of  stability  will  be  determined  by  assuming  13lr 
Ibs.  per  cubic  foot  for  the  weight  of  the  water.] 


288 


HYDROSTATICS. 


[379.1 


5.  The  space  between  the  pistons  H  and  A  is  filled  with 

water,  and  a  pressure  of  500  Ibs. 
is  exerted  on  the  small  piston 
by  means  of  a  lever  ;  if  the  dia- 
meter of  the  small  piston  is  1^ 
inches,  and  of  the  large  one  15 
E  inches,  what  will  be  the  pressure 
exerted  by  the  large  piston  ? 


FIG.  177. 


[This  machine  is  called  a  Hydraulic 
Press,  It  was  perfected  by  one  Bramah, 
who  packed  the  pistons  with  a  leather 
collar  in  such  a  way  that  the  pressure  of 

the  water  forced  the  leather  against  the  sides  of  the  cylinder,  thus 

keeping  the  pistons  water-tight.] 

EXERCISES. 

1.  In  Figs.  165  and  166,  will  the  surface  be  the  same  for  mercury  as 

for  water,  other  things  being  the  same  ? 

2.  Why  do  the  answers  to  Examples  1,  in  Articles  370  and  379,  differ  ? 

3.  If  a  vertical  square  be  entirely  submerged,  will  the  centre  of  pres- 

sure coincide  with  the  centre  of  gravity  of  the  square  ? 

4.  In  the  preceding  Exercise,  if  the  square  be  turned  about  a  horizontal 

line,  passing  through  its  centre  of  gravity,  will  the  depth  of  the 
centre  of  pressure  be  changed  ? 


CHAPTER 

HYDRODYNAMICS. 

380.  Mean  or  Average  Velocity.  —  The  velocity  is  not 
the  same  at  all  points  in  the  cross  section  of  a  stream, 
whether  the  stream  be  a  river,  or  canal,  or  in  a  pipe  or 
tube.  That  velocity  which,  being  multiplied  by  the  area 
of  the  cross  section,  will  equal  the  quantity  discharged  is 
called  the  mean  velocity. 

Let  Q  =  the  quantity  which  passes  a  section, 

S  =  the  area  of  the  cross  section,  and 

v  =  the  mean  velocity  ; 
then 


-- 
-  8' 

381.  Permanent  Flow.  —  If  the  same  quantity  passes 
all  the  transverse  sections  the  flow  is  said  to  \)Q  permanent  : 
otherwise  it  is  variable.     In  a  canal  the  flow  would  be  per- 
manent if  there  were  no  wastes  from  evaporation  or  leak- 
age ;  in  a  pipe  without  branches  the  flow  will  be  perma- 
nent throughout  its  length. 

382,  Variable  Velocities.  —  In   a  stream   of   variable 
sections  in  which  the  flow  is  permanent,  the  mean  veloci- 
ties are  inversely  proportional  to  the  transverse  sections  of 
the  stream. 

Let  v,  S,  and  Q  be  the  quantities  for  one  section, 

Vi,  Sij  and  Q  the  quantities  for  another  section; 
then,  according  to  Article  380,  we  have 
13 


290  HYDRODYNAMICS.  [383.1 


•  •          —    o  " 

v:      o 

383.  Problem. — Tojind  the  velocity  with  which  a  per- 
fect liquid  will  flow  through  an  orifice  in  the  base  of  a 
vessel. 

We  will  assume  that  the  particles  start  from  rest  at  a 
point  near  the  orifice,  and  that  the  velocity  of  their  exit 
is  produced  by  a  constant  pressure,  equal  to  the  weight  of 
the  water  vertically  over  them  (Article  367). 

Let  ab  be  the  short  distance  through  which  the 
velocity  is  generated, 

A~B  =  k,  the  height  of  the  liquid, 

TF",  the  weight  of  the  quantity  in  the  height  ab, 

$,  the  area  of  the  orifice, 

F,  the  pressure  of  the  liquid  above  the  orifice ; 
FIG.  178.  and 

v,  the  velocity  of  discharge. 

Then,  since  the  flow  is  supposed  to  be  without  resistance, 
the  conditions  are  essentially  the  same  as  Problems  1  and  2, 
pages  45  and  46  ;  hence 


= 


But  in  this  problem  we  have 
F= 

W= 
which  substituted  above  gives 

v  = 

which  is  the  same  as  that  of  a  body  falling  freely  through 
a  height  h  ;  see  Article  72. 


[384-386.]  DYNAMIC    HEAD.  291 

384.  The  velocity  through  an  orifice  in  the  side  of  a 
vessel  Mail  also  be 


in  which  h  is  the  depth  of  the  orifice  below  the  surface  ; 
fcr  the  pressure  against  the  side  is  the  same  as  the  vertical 
pressure  at  the  same  depth. 

385.  Head  due  to  a  Velocity.  —  The  preceding  equa- 
tion gives 


in  which  the  height  A,  corresponding  K>  the  velocity  v,  is 
called  the  head  due  to  ths  velocity,  or  simply  the  head. 
The  corresponding  velocity  is  called  the  velocity  due  to  the 
head. 

386.  Vertical  Pressure   on  the  top   of  a  Vessel.— 
If  a  piston  were  fitted  into  the  vessel  at  ,#,  Fig.  178,  and 
a  pressure  applied  to  it,  the  velocity  of  issue  would  be  in- 
creased.    To  find  the  resulting  velocity,  let 
P  =  the  pressure  on  the  surface, 
8  =  the  area  of  the  upper  surface, 
p  =  P  ~  $  =  the  pressure  on  a  unit  of  area, 
^  =  a  height  of  liquid  which  will  give  a  pressure 

equal  to  P, 

D  =  the  density  of  the  liquid,  and 
8  =  the  weight  of  a  unit  of  volume  of  the  liquid, 

=  Dg  (see  the  last  equation  of  Art.  85)  ; 
then  we  have 

P  = 


and  the  velocity  of  flow  will  be 
v= 


292 


HYDRODYNAMICS. 


[387-389.] 


This  is  called  finding  an  equivalent  head.  If  the  liquid 
issues  into  a  fluid  more  dense  than  air,  there  will  be  a 
counter-pressure.  If  h%  is  the  head  due  to  the  difference 
of  the  pressures  of  the  air  and  fluid,  then 


v  = 

387.  Pressure  of  the  Air. — When  a  vessel  is  in  the  air 
it  is  pressed  on  the  upper  surface  with  nearly  15  pounds 
to  the  square  inch,  which  is  equivalent  to  a  column  of 
water  of  the  same  base  and  about  34  feet  high.     If  a  ves- 
sel of  any  liquid  should  discharge  into  a  vacuum,  this  head 
must  be  added  to  the  head  of  the  liquid,  but  in  practice 
the  air  presses  against  the  issuing  stream  with  the  same 
pressure  per  unit  that  it  presses  against  the  top,  so  that  the 
head  due  to  the  pressure  of  the  air  is  not  considered. 

388.  Vertical  Jet. — If  the  issuing  jet  should  be  verti- 
cally upward,  as  in  Fig.  179,  and  there  were  no  resistances 


G  a 

FIG.  180. 

from,  the  air  or  the  sides  of  the  orifice,  the  jet  ought  k 
rise  as  high  as  the  free  surface  of  the  liquid  in  the  vessel. 
But  it  is  found  in  practice  that  it  always  falls  short  of 
that  height. 

389.  Orifices  in  the  Side  of  a  Vessel.— If  the  fluid 
issues  horizontally  from  an  orifice  in  the  side  of  a  vessel, 
the  jet  will  be  subjected  to  the  same  law  as  that  of  a  pro- 


[390-1.]         COEFFICIENTS    OF    DISCHARGE.  293 

jectile  thrown  horizontally.  Let  D,  Fig.  180,  be  an  orifice 
in  the  side  of  a  vessel,  and  DG  the  path  of  the  fluid  vein. 
In  Article  306  make  x  —  AG  and  y  —  DA,  and  we  have 


But  h  =  BD,  hence 


If  on  AB,  as  a  diameter,  a  semicircle  AHB  be  de- 
scribed, and  an  ordinate  DI  be  erected,  then,  from  geome- 
try, we  have 

hence,  the  range 


that  is,.  if  a  semicircle  be  constructed  on  AB  as  a  diame- 
ter, the  range  will  be  twice  the  ordinate  of  the  semicircle 
drawn  from,  the  orifice. 

Hence  the  maximum  range  AC  will  result  from  the 
flow  through  an  orifice,  E,  at  the  middle  of  the  height. 

Also,  the  ranges  from  two  orifices,  D  and  F,  equidistant 
from  the  centre,  E,  will  equal  each  other. 

390.  Oblique  Jet.  —  If  the  jet  be  oblique,  spouting  up- 
ward or  downward,  the  range  may  be  determined  by  the 
formulas  in  Article  299,  considering  the  vein  as  the  path 
of  a  projectile. 

Coefficients  of  Flow. 

391,  Coefficients  of  Contraction.  —  If  the  vein  issues 
through  a  thin  plate,  the  smallest  part  of  the  vein  will  be 
at  a  short  distance  from  the  orifice.     It  appears  that  the 
particles,  as  they  approach  and  issue  from  the  orifice,  tend 


294  HYDRODYNAMICS.  [392.., 

to  cross  each  other's  path,  and  by  thus  interfering  with 
each  other  first  produce  contraction  and  afterward  expan- 
sion, as  shown  in  Fig.  181. 

Let  /S  be  the  area  of  the  orifice  ab, 

Si,  the  area  of  the  contracted  part  cd,  and 
mi,  the  coefficient  of  contraction  ; 
then 

Si  =  m^S. 

For  a  very  thin  plate,  the  distance  of  smallest  section 
of  the  vein  from  the  orifice  will  be  equal  to  the  radius  of 
the  orifice,  and  the  diameter  of  the  smallest  section  will 
be  0.8  of  the  diameter  of  the  orifice,  and  the  coefficient  of 
contraction  will  be  the  square  of  0.8 ;  hence  for  a  thin 
plate  ml  =  0.64. 

For  an  adjutage,  that  is,  for  a  short  tube,  abdc,  whose 
length  is  two  or  three  times  the  diameter  of  the  orifice, 
attached  to  the  orifice,  the  fluid  vein  will 
just  fill  it,  and  the  coefficient  of  contraction 

will  be 

mi  =  1. 

In  Fig.  181,  if  a  conicaliy  convergent  tube 
form  the  adjutage,  the  convergent  part  being  of  the  form 
and  length  of  the  vena  contracta,  and  the  smallest  diame- 
ter of  the  tube  be  taken  for  the  orifice,  then 

392.  Coefficients  of  Velocity.— In  Fig.  179,  if  ^  be 
the  height  to  which  the  jet  will  rise,  then  will  the  velocity 
of  discharge  be 


but  the  theoretical  velocity  will  be 

v 
hence 


[393.]  COEFFICIENTS    OF    DISCHARGE.  295 


from  which  the  value  of  v^  may  be  determined. 

Or,  from  the  first  equation  of   Article  389,  we  ha\e 
(writing  v±  for  v), 


by  which  vt  may  be  computed. 

Let  mz  =•  —  =  the  coefficient  of  velocity,  then 


For  a  mere  orifice  in  a  thin  plate,  .     .     .     m^  =  0.98 
For  a  short  tube,  Fig.  182,    .....     mz  =  1.00 

393.  Coefficients  of  Discharge.  —  The  coefficient  of 
discharge  is  the  ratio  of  the  actual  discharge  to  that  of  the 
theoretical.  Let  the  quantity  which  flows  through  an 
orifice,  or  pipe,  or  stream  be  measured,  and  the  quantity 
which  should  flow  be  computed  ;  and  let 

Q  =  the  quantity  of  theoretical  flow, 
Qi  =  the  quantity  of  actual  flow,  and 
m  =  the  coefficient  of  discharge  ; 
then 

Ql  =  mQ. 

But  the  actual  flow  equals  the  mean  velocity  in  the  section 
considered  into  the  area  of  the  section  ;  hence,  from  Arti- 
cle 380  and  the  two  preceding  articles,  we  have 


21)6  HYDRODYNAMICS.  [394.] 

which,  compared  with  the  preceding  value  of  Q^  gives 


m  — 
From  this  we  find 


from  which  the  coefficient  of  the  mean  velocity  may  be 
found  from  the  coefficients  of  discharge  and  contraction. 
For  an  orifice  in  a  very  thin  plate,  .  .  .  m  =  0.62 
For  a  short  tube,  Fig.  182,  .....  m  —  0.82 
A  comparison  of  these  results  shows  that  the  effect  of 
the  short  tube  is  to  reduce  the  amount  of  contraction  (pro- 
vided there  is  one  in  the  tube),  but  that  the  interference 
of  the  particles  or  filaments  still  reduces  the  velocity  of 
discharge.  If  a  small  hole  be  made  in  the  side  of  the 
tube,  at  a  distance  from  the  inside  of  the  vessel  equal  to 
the  radius  of  the  orifice,  air  will  rush  into  the  tube,  show- 
ing that  there  is  a  negative  pressure  on  the  tube.  If  a 
pipe  be  attached  to  the  tube  so  as  to  cover  the  hole  and 
extend  down  into  a  vessel  of  water,  the  water  will  rise  in 
the  tube  to  balance  the  negative  pressure,  the  height, 
according  to  Article  385,  being  nearly  equal  to 


in  which  0.18  equals  1-0.82. 

394.  Large  Orifices.  —  To  find  the  mean  velocity  of  dis- 
charge through  a  large  orifice  in  the  base  of  a  vessel. 

If  the  orifice  is  so  large  compared  with  the  cross  section 
of  the  vessel  as  to  cause  a  perceptible  velocity  of  the  upper 
surface  of  the  liquid,  the  mean  velocity  of  discharge  may 
exceed  that  due  to  the  head  ;  for  all  the  particles  will  have 
an  initial  velocity  which  is  itself  equivalent  to  a  head  cor- 


[394.]  LARGE    ORIFICES.  297 

responding  to  that  velocity.     Therefore,  the  head  due  to 
the  discharge  will  be  the  head  of  the  liquid  in  the  vessel, 
plus  the  head  due  to  the  velocity  of  the  surface. 
Let  6'  be  the  section  of  the  orifice, 

Si,  that  of  the  vessel  at  the  surface  of  the  liquid, 
v  and  v-b  the  corresponding  velocities. 
Then,  according  to  Article  382  we  have 


and  the  head  due  to  this  velocity  will  be,  according  to 
Article  385, 


which,  added  to  the  head  h  of  the  liquid,  gives,  according 
to  Article  386, 


/ 

=  V 


from  which  we  readily  find 


If  s  =  Si,  v  =  oo  ;  that  is,  the  velocity  must  be  infinite 
in  order  that  the  section  of  the  issuing  vein  at  the  orifice 
shall  equal  that  at  the  surface  of  the  liquid. 

If  s  is  so  small  compared  with  Si  that  it  may  bo  neglected 
then  we  have 


as  found  in  Article  383. 
13* 


HYDRODYNAMICS.  [395.1 

395.  External  Pressures  Considered. 
Let  p  be  the  pressure  per  unit  of  area  on  the  issuing 
vein  due  to  the  atmosphere  or  other  fluid,  and 
p^  the  pressure  per  unit  of  area  on  the  upper  sur- 
face of  the  liquid ; 

then  the  head  due  to  the  difference  of  these  pressures  will 
be  (Article  386), 


which  must  be  added  to  the  head  of  the  liquid ;  hence 

* 


S? 

A  discussion  of  this  equation  will  give  several  of  the 
preceding  ones. 

[We  cannot  follow  these  modifications  further  in  an  elementary 
work,  but  will  add  that  the  formulas  have  been  founded  on  the 
hypothesis  that  the  velocity  of  the  particles  at  their  exit  was  gen- 
erated in  an  infinitesimal  of  space  (Article  383),  but  it  is  evident 
that,  in  a  perfect  fluid,  all  the  particles  in  the  vessel  will  be  put 
in  motion  as  soon  as  the  liquid  begins  to  flow.  If  the  vessel  be 
prismatic,  and  all  the  horizontal  sections  are  assumed  to  remain 
horizontal,  and  the  vessel  kept  constantly  full,  the  velocity  of  the 
particles  in  the  upper  surface  of  the  liquid  being  zero,  and  t  the 
time  for  a  particle  in  the  upper  surface  to  reach  the  orifice  ;  then 
it  is  found  by  means  of  the  Calculus  that  the  velocity  of  exit 
will  be 


e  +1 

in  which  v  has  the  value  given  in  the  preceding  equation,  and  «  i* 
the  base  of  the  Naperian  logarithms.] 


1396,  397.] 


FLOW   THROUGH  WEIRS. 


299 


396.  A  Weir  is  an  opening  in  the  side  of  a  vessel  for 
the  discharge  of  a  liquid,  in  which  the  upper  surface  of 
the  liquid  is  &free  surface. 

397.  A  Rectangular  Notch. — To  find  the  quantity  of 
liquid  which  will  flow  from-  a  rectangular  notch  in  the 
side  of  a  vessel,  the  vessel  being  kept  constantly  full. 


a  a 


FIG.  183. 


FIG.  184. 


Let  ABOD  be  the  notch,  and  x  the  distance  of  any  fila- 
ment from  the  surface  EC.  The  velocity  of  any  fillet 
will  be 


which  is  the  equation  of  a  parabola,  v  being  an  ordinate, 
a?  an  abscissa,  and  2g  the  parameter.  Let  h  —  AI>,  then 
the  velocity  of  the  liquid  at  A  will  be 


v  — 


In  Fig.  184,  take  AE=  V%gh,  and  construct  the  para- 
bola BE,  then  will  the  velocity  at  any  point  in  the  verti 
cal  AB  be  represented  by  an  ordinate  of  the  parabola 
drawn  through  that  point.  The  quantity  which  will  flow 
through  the  orifice  in  a  unit  of  time  will  be  represented 
by  the  area  of  the  parabola  ABE,  multiplied  by  the  width 
b  =  B(J,  Fig.  183  ;  hence  in  a  time  t  it  will  be  represent- 
ed by  the  area  of  the  parabola  multiplied  by  the  breadth 


300  HYDRODYNAMICS.  [398-401.] 

of  the  weir  and  by  the  time  t.  But  the  area  of  a  parabola 
is  two-thirds  the  area  of  its  circumscribed  rectangle  ;  hence 
we  have  for  the  quantity  discharged  in  the  time  £, 

Q  = 


398.  The  mean  velocity  of  the  discharge  through  a 
rectangular  notch  at  the  contracted  sect-ion  will  be 


mbht 

that  is,  two-thirds  of  the  maximum  velocity  through  the 
notch. 

399.  The  coefficient  of  discharge  over  a  weir  depends 
upon  the  head.     If  the  head  is  very  small  the  coefficient 
will  be  small ;  but  for  ordinary  cases  we  have 

m  =  0.62  nearly. 

400.  Flow  through  a  submerged  rectangular  ori- 
fice in  the  side  of  an  upright  vessel. 

Let  h  be  the  head  above  the  base  of  the  orifice, 
AI,  the  head  above  its  upper  end,  and 
b,  its  width. 

The  discharge  will  be  the  same  as  that  due  to  the  differ- 
ence of  two  weirs  having  the  same  breadth  b,  and  heads 
A  and  A,  respectively ;  hence  the  formula  of  Article  397 
becomes 

Q  =  %mbtV%g(h*-h*). 

401.  Flow  in  Long  Pipes. — The  law  of  resistance  in 
long  pipes  is  rather  assumed  than  deduced.     It  is  assumed 
in  regard  to  the  velocity,  that  the  resistance  due  to  the 
adhesion  of  the  liquid  to  the  sides  of  the  pipe  varies  us 


1401.] 


LONG  PIPES. 


301 


the  square  of  the  velocity,  and  that  due  to  viscosity  varies 
directly  as  the  velocity,  so  that  if  a  and  b  are  two  constants, 
the  law  will  be  expressed  by 


In  regard  to  the  dimensions  of  the  pipe  the  resistance  will 
vary  directly  as  the  length  and  also  as  the  contour  (or 
wetted  perimeter)  of  the  pipe,  and  it  is  also  assumed  to 
vary  inversely  as  the  area  of  the  cross  section  of  the 
stream. 

If  I  be  the  length  of  the  pipe,  S  the  cross  section,  and  o 
the  contour,  or  wetted  perimeter,  then  the  law  will  be  ex- 
pressed by 


or 


FIG.  185. 


in  which  V  equals  b  —•  a. 

THE  TOTAL  HEAD  WILL  BE  the  head  AD  =.  h,  of  the 
upper  reservoir,  plus  the  head  due  to  the  fall,  or  BD  sin  0, 
minics  the  head  in  the  lower  reservoir,  if  there  be  one. 
The  total  head,  minus  the  head  due  to  the  velocity  of  dis- 
charge, will  equal  the  total  resistance.  If  the  pipe  be 
prismatic  and  full,  the  velocity  will  be  uniform  through- 
out its  length. 

Let  II  =  the  total  head, 

I  =  DB,  the  length  of  the  pipe, 
tf>  =  the  inclination  of  the  pipe  to  the  horizontal, 
h  =  AD,  and  v  =  the  velocity  of  discharge ; 
then,  if  the  discharge  be  into  the  air,  we  have 


302  HYDRODYNAMICS.  [403.J 

H  =  h  -|-  I  sin  <J>, 
and 


Numerous  experiments  have  been  made  by  European 
ei.gineers  to  determine  the  constants  a  and  J',  among 
which  those  of  Prony,  Bossut,  and  Eytelwein  are  among 
the  most  noted.  According  to  the  results  of  these  experi- 
ments, D'Aubisson,  a  French  writer,  finally  wrote  the 
equation  as  follows  : 

H-$-  =  0.000104392  ^(^  +  0.180449*;). 

20r  # 

402.  Circular  Pipes.  —  The  section  of  pipes  being'  cir- 
cular, if  D  be  the  diameter,  we  have 


and 

c  =  irD, 
and  the  preceding  equation  becomes 

H—  0.015536v2  =  0.000417568  io*  +  0.1804490). 

If  the  quantity  of  discharge  be  given,  the  velocity  may  be 
eliminated. 

Let  Q  =  the  quantity  discharged,  then 


.:v  =  1.27324^; 

which,  substituted  in  the  preceding  equation,  gives 

*  For  a  history  of  this  and  other  formulae  pertaining  to  the  flow  of 
water  in  streams,  see  Report  on  the  Hydraulics  of  the  Mississippi  River, 
by  Humphreys  and  Abbott,  pp.  207  to  220. 


[403-405.]  FLOW  IN  EIVERS.  303 

H-  0.025187  ^  =  0.0006769  -^.(^+0.141724^^7). 

403.  To  find,  the  diameter  of  the  pipe  that  will  give 
a  given  discharge.  As  1)  is  involved  to  the  fifth  power, 
it  is  not  practicable  to  make  a  direct  solution.  Omitting 
the  terms  of  least  value,  that  is,  the  second  terms  in  each 
member,  and  we  have  the  following  approximate  value : 

D  -  0.2323 


which,  for  velocities  above  two  feet  per  second,  is  consid- 
ered sufficiently  accurate. 

404.  Condition  of  the   Pipe. — The   experiments  of 
Mons.  Darcy  showed  that  cast-iron  pipes,  whose  interior 
surface  was  covered  with  deposits,  offered  a  much  greater 
resistance  than  new  and  clean   cast-iron  ones,  and  that 
when  the  internal  surface  was  covered  with  bitumen,  or, 
in  other  words,  was  practically  polished,  the  resistance  was 
least. 

Bends  in  pipes  also  diminish  the  velocity,  and  sharp 
turns  offer  much  greater  resistance  than  rounded  ones. 

405.  Flow  in  Rivers  and  Canals. — The  formula  for 
the  flow  in  rivers  and  canals  is  of  the  same  general  form 
as  that  given  in  Article  401  for  the  flow  in  long  pipes, 
except  that  when  a  portion  only  of  the  length  of  the  stream 
is  considered,  and  the  mean  velocity  is  constant,  the  head 
due  to  the  terminal  velocity  will  be  neglected ;  for  the 
initial  velocity  will  be  the  same  as  the  terminal. 

Using  the  constants  which  were  determined  by  Prony 
for  this  case,  we  have 

H-  0.000111415      (^  +  0.217786*;); 


304-  HYDRODYNAMICS.  [406,407.] 

in  which  c  is  the  wetted  perimeter,  that  is  the  line  in  the 
cross-section  which  is  in  contact  with  the  water,  and  H  is 
the  fall  for  the  length  L 

Let  Q  —  v/S  =  the  quantity  of  flow,  and  substituting  in 
the  preceding  equation 

v  =  ®, 

and  omitting  the  last  term,  we  find 

£r=  94.738x5  A  K. 
y      co 

406.  Character  of  the  Bed  of  the  Stream. — Mons, 
Darcy  found  that  streams  having  cement  beds  offered  the 
le^st  resistance  to  the  motion,  and  that  the  resistance  in- 
creased in  the  order  of  the  following  substances :  Cement, 
planks,  bricks,  gravel,  and  coarse  pebbles.     (See  Morin's 
Hydraulique,  Troisieme  Ed.,  p.  147.) 

407.  Cross  Section  of  the  Stream. — It  is  found  by 
observation  that  the  surface  of  a  stream  is  not  horizontal 
in  its  cross  section,  but  that  it  is  highest  where  the  velocity 


Fio.  186. 


is  greatest.  This  is  accounted  for  by  the  fact  that  when 
the  fluid  is  in  motion  it  does  not  exert  as  great  a  side  pres- 
sure as  when  it  is  at  rest ;  and  as  the  velocity  near  the 
shore  is  very  small,  it  requires  a  greater  head  near  the 


L408.J 


BACKWATER. 


305 


middle  of  the  stream,  where  the  velocity  is  greatest,  to 
balance  the  pressure  at  the  sides.  It  may  also  be  observed 
that,  iu  order  to  produce  a  velocity,  there  will  be  a  greater 
pressure  in  the  direction  of  motion  in  order  to  overcome 
the  resistances  to  motion,  than  there  will  be  in  a  trans- 
verse direction. 

408.  Backwater  caused  by  a  Dam  in  a  Stream. — 

If  a  dam  be  made  across  a  stream,  or  partly  across  it,  so 
as  to  elevate  the  surface  at  the  place  of  the  dam,  the  sur- 
face of  the  water  above  the  dam  will  not  be  horizontal. 
If  a  horizontal  line  CK,  Fig.  187,  be  drawn  through  the 
crest  of  the  dam,  the  surface  of  the  water  in  the  pond  will 


PIG.  187. 


be  entirely  above  the  line,  the  difference  between  the 
surface  and  line  being  very  small  at  first,  but  increasing 
gradually  as  the  distance  from  the  dam  increases.  The 
natural  surface  may  also  be  elevated  for  a  long  distance 
back  of  the  point  K^  where  the  horizontal  through  C  in- 
tersects the  natural  surface  of  the  stream.  The  elevation 
of  the  surface  above  the  horizontal  CK,  including  also  the 


306  HYDRODYNAMICS.  L409.] 

elevation  back  of  the  point  If,  is  called  backwater,  and  is 
also  sometimes  called  a  remou. 

Backwater  is  caused  parti}7  by  the  inertia  of  the  liquid 
and  partly  by  its  viscosity.  As  the  stream  approaches  the 
dam  its  velocity  is  checked,  because  the  pressure  on  the 
front  side  of  a  particle  exceeds  that  on  the  rear  side,  and 
when  the  velocity  is  thus  reduced  the  particles  offer  a  re- 
sistance to  those  which  succede,  and  thus  the  resistance  is, 
so  to  speak,  extended  up  the  stream.  The  resistance  due 
to  viscosity  still  further  increases  this  effect. 

Fig.  187  shows  a  section  of  the  river  Weser,  in  Ger- 
many, at  the  place  of  a  certain  dam,  but  the  horizontal 
scale  is  much  less  than  the  vertical.  The  mean  width  of 
the  stream  was  354  feet,  the  mean  depth  about  2.47  feet, 
the  depth  of  the  water  just  above  the  dam  was  9.82  feet, 
and  hence  the  surface  was  raised  7.35  feet.  The  slope  of 
the  stream  was  quite  uniform  for  a  distance  of  ten  miles, 
and  averaged  0.000454  per  foot.  At  the  point  K,  where 
the  horizontal  CK,  through  the  crest  of  the  dam,  intersected 
the  natural  surface  of  the  stream,  it  was  found  by  actual 
measurement  that  the  surface  was  elevated  over  15  inches. 
The  distance  CK  was  over  three  miles.  At  a  distance  of 
four  miles  above  the  dam,  or  about  one  mile  above  the 
point  K,  the  elevation  of  the  surface  caused  by  the  dam 
was  about  nine  inches. 

In  ordinary  streams  the  width,  depth,  and  the  character 
of  the  bed  are  such  variable  quantities  that  the  extent  of  tho 
backwater  cannot  be  very  accurately  computed  on  a  theo- 
retical basis,  but  empyrical  formulas  have  been  given 
which  will  give  an  approximate  result  when  applied  to 
rivers  of  about  the  dimensions  of  the  Weser.  (See  D' Au- 
bisson's  Hydraulics,  Article  166.) 

409.  Back-water  in  Rapid  Streams. — If  the  stream  is 


[409.]  .         PROBLEMS.  307 

very  rapid,  or  rapid  compared  with  its  depth,  the  remous 
will  be  modified,  presenting  an  appearance  similar  to  that 
shown  in  Fig.  188.  In  this  case  there  is  a  comparatively 
sudden  change  in  the  velocity  at  the  head  of  the  pond. 


Problems. 

\ .  A  prismatic  vessel  is  kept  constantly  full  of  a  liquid; 
required  the  time  of  discharging  a  given  quantity  through 
a  small  orifice  in  the  base. 

Let  Q  be  the  quantity,  A,  the  height  of  the  liquid,  s, 
the  area  of  the  orifice,  and  t  the  time ;  then 

Q  =  msvt 


From  this  we  find  the  time  of  discharge  to  be 

Q 


2.  determine  the  time  in  which  a  prismatic  vessel  will 
empty  itself  by  an  orifice  in  the  base. 

Let  S  be  the  area  of  the  free  surface  of  the  liquid,  s 
that  of  the  orifice,  x  the  variable  head,  v  the  velocity  of 
discharge,  and  V  the  velocity  of  descent  of  the  free  sur- 
face of  the  liquid.  Then 


308  HYDRODYNAMICS.  [409.] 


8  S 


which,  compared  with  equation  (3),  page  34,  shows  that 
the  law  of  descent  of  the  surface  is  the  same  as  that  of 
falling  bodies,  or  rather,  it  is  the  same  as  that  of  a  body 
projected  vertically  upward  ;  and  from  equation  (4)  on 
page  34  we  have 

_  2A 
t  —  —  , 

v  ' 

in  which  h  is  the  height  to  which  a  body  will  rise  when 
projected  vertically  upward  with  a  velocity  v.  Were  the 
velocity  to  remain  uniform  from  the  instant  that  it  is  pro- 
jected, the  time  required  to  go  the  same  distance  would  be 


.-.  t  =  2^  ; 

hence  the  time  required  for  the  vessel  to  empty  itself  will 
be  twice  that  required  to  discharge  the  same  quantity  when 
the  vessel  is  kept  constantly  full.  If  Q  be  the  contents 
of  the  vessel,  we  have,  by  multiplying  the  last  equation  of 
the  preceding  article  by  two, 


3.  To  determine  the  form  of  a  vessel  such  that  the  free 
surface  of  the  liquid  shall  descend  uniformly  as  it  dis- 
charges itself  through  a  small  orifice  at  the  lower  end  of 
the  vessel. 

Let  the  vessel  be  one  of  revolution,  h  the  height  of  the 


[409.]  PROBLEMS.  309 

upper  base  above  the  orifice,  r  the  radius  of  the  upper 
base,  x  the  height  of  any  section  above  the  orifice,  y  its 
radius,  and  Ax  the  thickness  of  the  horizontal  laminse  ; 
then  the  conditions  of  the  problem  require  that  the  times 
of  descent  through  the  small  distance  Ax  shall  be  the  same 
ior  all  positions  of  the  upper  surface.  Let  *  be  the  area 
of  the  orifice,  and  t  the  time  of  discharging  a  quantity 
equal  to  any  lamina,  then 

msV^gh.t  =  Trr*.  Ax, 
and 

msV^gx.t  =  Try*.  Ax  ; 

and  dividing  one  of  these  equations  by  the  other  and  re- 
ducing, gives 

h 


which  is  the  equation  of  a  biquadratic  parabola.  Clep- 
sydras or  Water  Clocks  involve  these  principles.  If  the 
time  jTof  the  complete  discharge  of  the  vessel  be  given, 
then  will 

A  _  Ax  _  msVSgh  __ 

~T=    T=  TT/"2  ''C' 

in  which  c  is  a  constant  and  is  the  rate  of  discharge.  From 
this  equation  we  find 
A 


and  the  equation  of  the  curve  becomes 

y=\/ 


310  HYDRODYNAMICS.  [409.] 

Making  x  =  h,  we  have,  for  the  radius  of  the  upper  base 
of  the  vessel, 


y    r    y 


EXAMPLES. 

1 .  A  cylindrical  vessel,  whose  height  is  3  feet,  radius  of  the 

base  6  inches,  is  filled  with  water,  and  discharges 
itself  through  an  orifice  in  the  base  ;  if  the  diameter 
of  the  orifice  is  one-half  of  an  inch,  and  the  coefficient 
of  discharge,  w,  is  0.62,  in  what  time  will  the  vessel 
empty  itself? 

2.  What  quantity  of  water  will  flow  over  a  weir  whose 

breadth  is  2  feet  and  constant  depth  10  inches,  in  45 
minutes  ? 

3.  It  is  required  to  construct  a  water  clock  which  will 

empty  itself  in  10  minutes,  the  surface  descending 
uniformly.  The  height  of  the  vessel  being  24  inches 
and  radius  of  its  upper  base  3  inches  ;  required  the 
equation  of  the  curve,  and  the  area  of  the  orifice,  the 
coefficient  of  discharge  being  0.62. 

4.  What  quantity  of  water  will,  in  one  hour,  flow  through 

a  pipe  1,500  feet  long,  2  inches  in  diameter,  the 
open  eiid  being  25  feet  below  the  level  of  the  reser- 
voir? 

EXEKOI8ES. 

1.  If  a  vessel  having  an  orifice  in  its  base  be  filled  successively  with 

alcohol,  water,  and  mercury,  which  will  require  the  least  time  to 
empty  itself,  considering  each  liquid  as  perfect  f 

2.  If  a  cylindrical  vessel  be  half  filled  with  mercury  and  the  remaining 

half  with  water,  will  the  velocity  of  discharge  of  the  mercury 


[409.]  EXERCISES.  31J 

through  an  orifice  in  the  base  of  the  vessel  be  the  same  as  if  the 
vessel  were  entirely  filled  with  mercury  ? 

3.  If  the  lower  half  is  water  and  the  upper  half  mercury,  will  the  flow 

be  the  same  as  in  the  preceding  exercise  ? 

4.  If  two  liquids  of  different  densities  are  thoroughly  mixed,  will  the 

velocity  of  flow  from  an  orifice  in  a  vessel  be  the  same  as  for  each 
liquid  separately  ? 

5.  A  block  is  floating  on  the  water  in  a  vessel,  when  an  opening  is  sud- 

denly made  in  the  base  of  the  vessel ;  considering  that  the  surface 
of  the  water  falls  with  a  decreasing  acceleration,  will  the  depth  of 
flotation  of  the  block,  during  the  discharge  of  the  water,  be  the 
same  as  before  the  discharge  began  ?  Will  the  depth  of  flotation 
remain  constant  during  the  discharge  ? 

6.  If,  in  a  pond  which  receives  no  supply,  an  opening  is  made  in  one 

side  so  that  the  water  will  flow  out,  will  the  surface  remain  at  a 
true  level — that  is,  parallel  to  the  surface  as  it  stood  before  the 
opening  was  made  ? 

7.  If  a  vessel  filled  with  water,  and  having  an  orifice  near  its  bottom,  is 

placed  on  a  platform  and  made  to  ascend  with  a  uniform  accelera- 
tion, will  the  velocity  of  flow  through  the  orifice  be  the  same  as  if 
the  vessel  were  at  rest  ? 

8.  In  Fig.  166,  if  there  be  an  orifice  in  the  base  near  the  outer  edge, 

will  the  velocity  of  discharge  be  the  same  when  the  vessel  ia 
rotating  as  when  it  is  at  rest  ? 

9.  When  water  is  flowing  uniformly  in  a  pipe  is  the  pressure  against  th« 

sides  of  the  pipe  the  same  as  if  the  discharge  be  stopped  ? 


CHAPTER  XXII. 


GASES   AND   VAPORS. 

410.  A  Gas  is  a  fluid  whose  particles  are  in  a  constant 
state  of  repulsion.     Common  air  is  taken  as  a  type  of  gases. 

411.  Pressure  of  the  Atmosphere. — If  a  tube,  32  or 
33  inches  long,  be  closed  at  one  end  and  filled  with  mer- 
cury, and  the  open  end  be  closed  with  the  finger  until  the 
tube  is  inverted  and  the  end  submerged  in  a  vessel  of  mer- 
cury, then  if  the  finger  be  removed  the  mercury  in  the 

tube  will  fall  to  some  point 
B,  and  remain  nearl}7  sta- 
tionary. The  column  AB 
is  sustained  by  the  pressure 
of  the  atmosphere  upon  the 
surface  of  the  mercury  in 
the  vessel  A  ;  and  hence  the 
weight  of  a  column  of  mer- 
cury equal  to  AB^  having  a 
square  inch  for  its  base,  will 
equal  the  pressure  of  the 
atmosphere  upon  a  square 
inch  of  surface.  The  aver- 
age height  of  the  column 
AB  at  the  level  of  the  sea 
is  about  29.92  inches  (say  30 
inches,  or  760  millimeters) 
of  mercury,  or  about  34  feet  of  water.  Hence  the  mean 
pressure  of  the  atmosphere,  at  the  level  of  the  sea,  is  about 


FIG.  189. 


[412-414.]  MARIOTTE'S  LAW.  313 

14.7  pounds  (say  15  pounds)  per  square  inch.  This  is  called 
the  pressure  of  one  atmosphere.  The  pressure  of  the  at- 
mosphere diminishes  as  the  distance  above  the  earth  in- 
creases, and  increases  for  depths  below  the  surface. 

412.  The  Barometer. — If  the  tube  and  vessel  shown  in 
Fig.  189  be  encased  in  such  a  way  as  to  retain  their  rela- 
tive positions  while  they  are  carried  from  place  to  place, 
and  the  tube  be  provided  with  a  suitable  scale  for  reading 
the  height  of  the  end  B  of  the  mercurial  column,  above 
the  surface  A,  the  instrument  is  called  a  barometer.     By 
means  of  it  the  pressure  of  the  atmosphere  may  be  readily 
determined  for  any  place  and  at  any  time.     There  are 
numerous  modifications  in  the  details  of  different  barome- 
ters which  are  explained  in  descriptive  works  upon  the 
subject. 

413.  Height    of   a   Homogeneous    Atmosphere. — 
If  the  atmosphere  were  of  uniform  density,  and  the  same 
as  that  at  the  level  of  the  sea,  its  height  would  be  found 
by  multiplying  30  inches  (the  height  of  the  mercurial 
column)  by  the  ratio  of  the  density  of  mercury  to  that  of 
air.     Mercury  is  about  13£  times  as  dense  as  water,  and 
water  773  times  as  dense  as  air  when  the  barometer  stands 
at  30  inches  ;  hence  the  height  would  be 

2fc  x  773  x  13£  =  26088  feet  nearly  =  5  miles  nearly. 
But  the  actual  height  is  supposed  to  be  from  50  to  100 
miles.     This  is  determined  by  its  effect  in  deflecting  the 
rays  of  the  sun. 

414.  Boyle's  (or  Mariotte's)  Law.— Boyle  in  England 
and  Mariotte  in  France,  independently  of  each  other,* 
demonstrated  the  following  law :  

*  Writers  differ  in  regard  to  the  respective  dates  of  the  discovery. 
While  some  state  that  both  made  the  discovery  at  about  1668,  others 
give  to  Boyle  a  precedence  of  several  years  over  Mariotte. 
14 


314  GASES    AND    VAPORS.  [414.1 

For  the  same  temperature  the  density  of  a  gas  is  directly 

proportional  to  its  pressure. 
Both  these  discoverers  proved  this  ]aw  by  means  of  a 

tube,  called  Mariotte's  tube,  Fig.  190.  Mercury  was 
poured  into  the  tube  until  the  air  passage  from 
the  short  to  the  long  tube  was  just  cut  off.  This 
point  was  marked  zero,  and  the  pressure  of  the 
air  in  the  short  tube  was  that  of  one  atmosphere 
when  the  mercury  stood  at  this  point.  Mercury 
was  then  poured  into  the  long  tube  until  the  air 
in  the  short  column  was  compressed  to  one-half 
its  length,  when  it  was  found  that  the  upper  end 
[«,  of  the  long  column  was  about  30  inches  above 
the  upper  end  of  the  mercury  in  the  short  tube. 

Again,  filling  the  long  tube  until  the  air  in  the 
short  tube  is  compressed  to  one-quarter  of  its 
length,  it  will  be  found  that  the  column  in  the 
long  tube  above  that  in  the  short  tube  will  be 
twice  its  former  length,  and  so  on,  observing  in 
each  case  that  the  temperature  must  be  the  same. 
The  quantity  of  air  being  constant,  the  density 

will  be  inversely  as  the  volumes,  or  directly  as  the  pres- 

sure. 

Let  F"and  Ft  be  the  volumes  corresponding  to  pressures 

p  and  pi  per  square  inch,  and  D  and  Dl  their  densities. 

then 

L  —a  -  ^ 

Ft  ~p        D 


p 

If  Fi  and  p^  are  known,  the  volume  V  may  be  found  for 


[415,  416.] 


MANOMETERS. 


315 


any  pressure  p.     Considering  V  and  p  as  variables,  the 
preceding  equation  will  be  of  the  form 

ocy  =  ra, 

which  is  the  equation  of  an  hyperbola  referred  to  its 
asymptotes. 

415.  Boyle's  Law  is  only  approximately  correct. — 

For  many  years  after  the  announcement  of  Boyle's  law  it 
was  confirmed  by  different  experimenters,  and  the  law  dur- 
ing that  time  was  supposed  to  be  rigorously  correct,  but 
more  recently  the  more  precise  experiments  of  Despertz 
and  Regnault  have  shown  that  it  differs  for  different  gases 
and  is  not  rigidly  true  for  any  gas.  But  the  departure 
from  the  law  is  so  slight  that,  for  ordinary 
purposes,  it  may  safely  be  considered  as 
exact. 

416.  Manometers  are  instruments  for 
measuring  the  tension  of  a  gas.     The  ten- 
sion is  the  pressure  per  square  inch,  and  is 
often  compared  with  the  pressure  of  one 
atmosphere.     The  principle  of  the  mano- 
meter is  founded  on  Boyle's  law  of  the  com- 
pressibility of  gases.    There  are  many  kinds, 
but  the  closed  manometer,  Fig.  191,  is  one 
of  the  most  common.     It  consists  of  a  ver- 
tical tube  closed  at  the  upper  end,  the  lower 
end  opening  into  a  vessel  of  mercury  or 
other  liquid.      Another  tube   connects  the 
liquid  with  the  vessel  containing  the  gas,  so 
that  the  pressure  of  the  gas  when  it  is  ad- 
rnitted  into  this  tube  will  force  the  liquid  up 

the  long  one,  thereby  compressing  the  air  above  the  liquid. 
The  vertical  tube  is  provided  with  a  scale  for  indicating 


316  GASES    AND    VAPORS.  [417-419.] 

the  pounds  of  pressure  per  square  inch,  or  the  number  of 
atmospheres,  or  both,  as  may  be  desired. 

417.  Expansion  of  Gases  due  to  a  Change  of  Tem- 
perature.— Reliable  experiments  show  that  the  expansion 
of  all  gases  under  constant  pressure  may  be  considered  as 
the  same  for  each  degree  of  increase  of  temperature  for 
all  ranges  of  temperature.     Still  the  delicate  experiments 
made  by  Regnault  show  that  the  expansions  are  not  identi- 
cally the  same,  and  that  the  increase  of  volume  increases 
somewhat  more  rapidly  than  the  increase  of  temperature. 

Assuming  that  Mariotte's  law  is  rigorously  exact,  and 
that  the  rate  of  expansion  of  a  gas  is  the  same  as  that  of 
the  increase  of  its  temperature,  it  follows  that  the  tension 
of  a  gas  under  constant  volume  varies  directly  as  the 
change  of  temperature. 

418.  Coefficient  of  Expansion  due  to  Temperature. 
— The  volume  of  dry  air  under  a  constant  pressure  in- 
creases 0.002039  of  its  original  volume  for  each  increase 
of  1°  Fahrenheit,  and  this  is  called  the  coefficient  of  ex- 
pansion.    It  is  also  considered  as  the  coefficient  of  tension 
under  a  constant  volume  for  each  1°  Fahrenheit. 

Let  a  be  the  coefficient  of  expansion  (or  tension),  then 

a  =  0.002039  for  1°  F. 
=  0.003670  for  1°  C. 

419.  To  find  the  Volume  of  a  Gas  due  to  a  change 
of  Temperature  and  Pressure. 

Let  VO,_PO,  ^o?  °e  the  initial  volume,  tension,  and  temper 
ature,  and  V-^p^  t^  the  corresponding  terminal  values 
Then  will  the  change  of  temperature  be 

t\      ^oj 
and  hence  the  volume  due  to  this  change  will  be 


[420.J  EXPANSION  OF  GASES.  317 


and,  according  to  Mariotte's  law,  if  there  be  a  change  of 
pressure  at  the  same  time  the  volume  will  be 


Let  F^,  Pfr  t2,  be  the  quantities  for  another  pressure  and 
temperature;  then 


Dividing  the  former  equation  by  the  latter,  gives 
Ft 


420.  Perfect  Gas.  —  The  more  rare  a  gas  is  the  more 
perfect  it  is  considered  to  be.  A  perfect  gas  is  defined  to 
be  one  which  is  destitute  of  mass.  A  perfect  gas  does  not 
exist,  but  this  ideal  gas  serves  as  a  standard  of  comparison 
for  different  gases.  It  may  be  defined  to  be  the  limit 
towards  which  gases  approach  as  they  are  expanded  indefi- 
nitely. The  limit  towards  which  the  coefficient  of  expan- 
sion approaches  is  not  definitely  known,  but  it  is  assumed 
by  Rankine  to  be 

a  =  0.0020275  =  —  L-  for  1°  F. 


=  0.00365  =        forl°0. 


and  these  are  called  the  coefficients  of  expansion  for  a 
perfect  gas. 


318  GASES    AND    VAPORS.  [431.J 

Substituting  the  former  of  these  in  the  formula  above, 
and  taking  the  initial  temperature  at  the  melting  point  of 
ice,  in  which  case  t0  =  32°  F.,  we  have 


If  ti  were  taken  at  461.2°  F.  below  zero,  this  expression 
would  vanish.  The  ideal  temperature  —461.2°  F.,  or 
—  493.2°  F.  below  melting  ice,  is  called  the  ABSOLUTE  ZERO. 
This  temperature  cannot  be  even  approximately  reached 
by  any  known  process  ;  but,  according  to  the  formula,  it 
is  a  temperature  at  which  a  perfect  gas  would  lose  all  ex- 
pansive power. 

The  absolute  zero  is  used  because  the  formulas  for  the 
expansion  of  gases  due  to  temperature  are  simplified  when 
the  temperature  is  reckoned  from  that  point. 

421.  Thermometers  are  instruments  for  measuring 
changes  of  temperature.  They  are  made  on  the  principle 
of  the  uniform  rate  of  expansion  of  liquids.  Mercury  is 
most  commonly  used,  but  alcohol  is  sometimes  used  ;  and 
the  latter  is  necessarily  used  instead  of  mercury  for  tem- 
peratures below  —40°  F.,  for  mercury  freezes  at  about 
that  temperature.  Three  scales  are  used,  viz.  :  Fahren- 
heit's, marked  F.,  the  Centigrade,  marked  C.,  and  Eeau- 
mur's,  marked  R. 

Fahrenheit  chose  for  the  zero  of  his  scale  the  height  of 
the  mercury,  which  was  produced  by  a  mixture  of  salt  and 
ice.  This  he  believed  to  be  the  absolute  zero  of  cold.  The 
height  produced  by  the  boiling  point  of  water  he  marked 
212,  and  divided  the  space  between  these  points  into  equal 
parts.  The  freezing  point  of  water  was  32  divisions  above 
zero. 

The  zero  of  the  Centigrade  thermometer  is  at  the  frees- 


[422,  423.J          THERMOMETEIC   SCALES.  319 

ing  point  of  water,  and  the  boiling  point  of  water  is  marked 
100,  and  the  space  between  is  divided  into  equal  parts. 

In  Reaumur's  scale  the  zero  is  fixed  at  the  freezing 
point  of  water,  and  the  boiling  point  of  water  is  marked 
80,  the  space  between  being  divided  into  80  equal  parts. 

The  melting  point  of  ice  is  now  used  instead  of  the 
freezing  point  of  water,  for  the  latter  is  not  constant.  In 
all  thermometers  the  divisions  below  zero  are  considered 
minus. 

422.  To  Convert  one    Thermometric    Scale    into 
another. — The  number  of  divisions  between  the  melting 
point  of  ice  and  the  boiling  point  of  water  on  the  three 
scales  is 

F.  C.  R. 

180,  100,  80 ; 

or  as 

i  ?          i 

9'  9' 

But  the  Fahrenheit  scale  begins  32°  below  the  others ; 
hence,  if  F°,  C°,  and  R°  represent  the  degrees  on  the 
respective  scales  for  the  same  temperature,  we  have 

C°=  |  (F°-320) ;          R°=  5  (F°-32°) ; 
9  " 

from  which  we  find 

F°=:?C0-t-320,      F0=rR0  +  32°,      andC°=5R°. 
54  4 

423.  Compressed  Air. — Air,  compressed   to  several 
atmospheres,  may  be  used  instead  of  steam  for  driving 
engines.     If  the  engine  is  at  a  great  distance  from  the 
compressing  machine  it  is  a  more  desirable  power  than 


320  GASES    AND    VAPORS.  [433.] 

steam,  for  steam  will  condense  and --thus  lose  its  powerj 
but  compressed  air  may  be  conducted  for  miles,  if  neces- 
sary, without  any  loss  of  power  except  that  due  to  the 
leakage  of  the  pipes  in  which  it  is  conducted.  It  is  especi- 
ally serviceable  for  driving  engines  under  ground.  It  is 
the  chief  power  for  driving  engines  in  the  construction  of 
large  tunnels  and  in  underground  mining.  It  was  thus 
used  in  the  construction  of  the  Mount  Cenis  tunnel  in 
Europe,  and  the  Hoosac  tunnel  in  this  country. 

When  air  is  suddenly  compressed  heat  is  developed,  and 
sometimes  the  heat  becomes  so  intense  as  to  make  the 
compressor  very  hot.  If  this  heat  is  lost  while  the  air 
is  passing  from  the  compressor  to  the  engine,  power  is 
lost.  Methods,  therefore,  have  been  devised  for  keeping 
the  air  as  cool  as  possible  during  compression.  The  most 
practical  way  is  to  inject  a  spray  of  water  into  the  com- 
pressor while  the  air  is  being  compressed.  If  the  heat 
that  is  in  the  air  when  it  leaves  the  compressor  could  be 
maintained  without  extra  cost,  until  the  air  is  used,  there 
would  be  no  loss  due  to  the  development  of  the  heat,  and 
as  the  air  thus  heated  has  a  greater  tension,  it  would  be 
undesirable  to  reduce  the  temperature  during  compression 
BO  far  as  this  cause  is  concerned. 

[The  following  formula  gives  the  relation  between  the  pres- 
sure, density,  and  temperature  of  air  during  rapid  changes  of 
motion.  (See  Rankine's  Applied  Mechanics,  p.  582) : 


in  which  TJ  is  the  absolute  temperature  of  the  air  (or  gas)  when 
the  pressure  is^h  and  density  3i,  and  r2,  ^2,  82,  the  corresponding 
quantities  in  another  condition  of  the  gas.  7  is  the  ratio  of  spe- 


[424.J 


STEAM   OR  VAPOR. 


cific  heat  of  constant  pressure  to  the  specific  heat  of  constant 
volume,  which  for  a  perfect  gas  is  1.408,  which  value  is  suffi- 
ciently exact  for  dry  air.  The  quantity  of  air  being  constant,  the 
volumes  will  be  inversely  as  the  densities  ;  hence  we  have 


— =/— V ' 

T,   ~~\fl9  / 


Let  the  volume  Vi,  at  14.7  pounds,  be  100,  and  the  tempera- 
ture 60°  F. ,  then  will  the  absolute  temperature  be  TJ  =  521.2°. 
If  the  air  be  suddenly  compressed,  we  will  have,  according  to 
this  formula : 


Pressure,  Ibs.  per 
square  inch. 

Resulting  temper- 
atures. 
Deg.  P. 

Resulting 
Volumes. 
(Relative.) 

Relative  Volumes 
under  constant 
temperature. 
(Boyle's  Law.) 

14.7 

60 

100 

100 

30.0 

180 

60 

49 

45.0 

259 

45 

32 

60.0 

322 

37 

25 

75.0 

374 

31 

20 

90.0 

420 

28 

16 

105.0 

460 

25 

14 

After  passing  the  compressor,  if  the  heat  escapes  so  as  to  re- 
duce the  temperature  to  60°,  the  volumes  will  be  reduced  from 
those  in  the  third  column  to  those  in  the  fourth,  and  there  will 
be  a  corresponding  loss  of  tension  in  the  air.  (For  a  mathemat- 
ical computation  for  the  work  lost  in  using  compressed  air,  see 
Engineering  and  Mining  Journal,  July,  1873.)] 

424.  Steam  or  other  vapor,  separated  from  liquids,  may 
follow  the  same  laws  in  regard  to  expansion,  temperature, 
and  density,  as  air  and  other  gases.  If,  however,  the  steam 
be  in  contact  with  the  water  from  which  it  is  generated, 
the  temperature  cannot  be  increased  without,  at  the  same 
time,  increasing  both  the  tension  and  density  of  the  steam. 
14* 


322  GASES    AND    VAPORS.  [434] 

Steam,  in  this  condition,  is,  for  a  given  temperature,  al- 
ways at  its  maximum  tension,  and  also  at  its  maximum 
density.  It  may  be  said  to  be  constantly  at  its  dew  point. 
Steam  in  which  both  the  density  and  tension  change  on 
account  of  a  change  of  temperature  is  called  saturated  ^ 
steam.  But  steam  in  which  the  tension  may  change  with 
the  temperature  without  changing  its  density  is  called 
steam  gas.  It  follows  the  laws  of  permanent  gases  in  this 
respect.  The  tension  of  such  steam  may  greatly  exceed 
that  of  saturated  steam  for  high  temperatures,  and  when 
thus  heated  it  is  called  Superheated  Steam. 

Problems. 

1.  To  find  the  weight  of  a  cubic  foot  of  air  at  any  tem- 
perature and  pressure. 

The  weight  of  a  cubic  foot  of  air  at  32°  F.,  when  the 
barometer  is  at  30  inches,  is  0.08072  pounds  avoirdupois. 
For  a  given  mass  of  air  the  weight  of  a  given  volume  will 
be  inversely  as  the  temperature  and  directly  as  the  pres- 
sure, or  generally,  inversely  as  the  total  volumes  of  the  air 
under  the  different  conditions.  Hence,  the  weight,  TFi,  of 
a  cubic  foot  will  become,  according  to  Article  410, 


_ 
W0      Vi       1+  0.002039(£-  32°)  '  30 

w  _          0.08072^ 

1  ~~  28.0425  +  0.061170*  ' 
in  which  p±  is  the  reading  of  the  barometer. 

2.  Find  the  weight  of  a  cubic  foot  of  steam  at  any  tem- 
perature and  pressure. 

The  density  of  steam  is  five-eighths  of  the  density  of  air 
for  the  same  tension  and  temperature  ;  hence,  when  the 
pressure  is  given  in  pounds  per  square  inch,  we  have 


[424.] 


PEOBLEMS. 


323 


$  x  0.08072 


1  +  0.002039(2-32°)  '  14.75 

_  0.00342^ 

"13.7877+0.0300752' 

3.  A  spherical  air-bubble  rises  vertically  from  a  depth 
h  to  the  surface  of  the  liquid  /  required  the  equation  of 
the  curve  described  by  the  extremities  of  a  horizontal 
diameter. 

Let  r  =  the  radius  at  the  surface, 
y  =  the  radius  at  a  depth  x. 

The  volumes  will  be  inversely  as  the  pressures.  At  the 
surface  the  pressure  per  unit  is  represented  by  a  column 
of  water  34  feet  high,  and  at  a  depth  x  by  34+aj;  hence, 

•f-Try3: :  34+«  :  34; 
34T5 


or 


4.  Required  the  degree  of  exhaustion  from  the  receive? 
of  an  air-pump. 

At   each   stroke   of   the  piston    a      |  f    ^ 

quantity  of  air  is  removed  from  the 
receiver  R  and  the  pipe  a  c  equal  to 
the  volume  in  either  barrel  £  or  B'. 

Let  V  be  the  volume  in  the  re- 
ceiver and  pipes,  v,  the  contents  of 
either  barrel,  Z>0,  the  initial  density, 

Di,D2 Dn,  the  densities  after 

the  successive  strokes. 


X  =  d* 1— 5  — 


A 

P 

h 

R 

V 

V 

P\ 

\ 

i    . 

Flo.  193. 

324:  GASES  AND  VAPORS.  [424.J 

The  quantity  remaining  after  the  first  stroke  will  be 


which  will  be  uniformly  distributed  throughout  the  re- 
ceiver and  pipes,  and  the  density  will  be  reduced  to  D±  ; 
hence 


Similarly,  at  the  end  of  the  second  stroke,  we  have 

A  =  A(i-^)  = 

and  at  the  end  of  the  n"1  stroke 


The  density  Dn  cannot  be  zero  so  long  as  n  is  finite  ; 
hence  the  exhaustion  can  never  be  perfect.  Indeed,  the 
degree  of  exhaustion  falls  far  short  of  that  indicated  by  the 
formula,  for  the  valves  and  pistons  cannot  be  made  per- 
fectly air-tight,  and  it  requires  some  pressure  to  operate 
the  valves,  so  that  after  a  few  strokes  the  exhaustion,  in 
practice,  proceeds  very  slowly. 

5.   To  determine  elevations  by  means  of  a  barometer. 

In  order  to  solve  this  problem  it  is  necessary  to  know 
the  law  of  diminution  of  the  pressure  of  the  atmosphere. 
Consider  three  consecutive  strata  of  the  atmosphere  of 
equal  thicknesses,  but  so  thin  that  the  density  of  each 
may  be  considered  uniform.  The  pressure  on  the  top  of 
the  upper  stratum  "will  be  the  weight  of  the  atmosphere 
above  it,  which,  if  it  is  near  the  level  of  the  sea,  will  be 
nearly  14.75  pounds.  Let  this  pressure  be  p&  The  pres- 


[424.]  PROBLEMS.  325 

sure  upon  the  second  stratum  from  the  top  will  be^?l5 
plus  the  weight  of  the  stratum  ;  hence, 

Pi  —flo  =  the  weight  of  the  first  stratum. 
Similarly, 

Pi  —  P\  =  the  weight  of  the  second  stratum. 
The  weights  of  the  strata  are  as  their  densities,  or 


But,  according  to  Mariotte's  law,  the  densities  vary  directly 
as  the  pressures  ;  hence, 

D0:  Di\\ps'.\\ 


These  quantities  follow  the  law  of  a  geometrical  pro- 
gression ;*  hence,  the  natural  numbers,  1,  2,  3,  etc.,  which 
are  as  the  numbers  of  the  successive  laminae,  will  be  the 
logarithms  of  the  successive  pressures.  But  the  system  of 
logarithms  remains  to  be  determined.  The  thicknesses  of 
the  laminae  being  constant,  the  distance  below  the  initial 
point  will  equal  the  thickness  of  a  lamina  into  the  number 
of  laminae. 

*  Let  a  be.  the  first  term  of  a  geometrical  series, 
r,  the  ratio,  and 
n,  the  nth  term  ; 
then  for  three  consecutive  terms  we  have 


then  if  arn  be  substituted  for  j90,  arn+lfozpi,  and  ar"*2  for  pt,  the 
expression  in  the  text  will  reduce  to  the  identical  equation 


326  GASES    AND    VAPORS.  [434.1 

Let  Aa  =  the  thickness  of  a  lamina, 
z  =  the  number  of  laminae, 
h  =  z.Aa  =  the  distance  from  the  initial  point, 
p0  =  the  pressure  at  the  initial  point, 
p  =  the  pressure  at  the  distance  h  from  the  initial 

point, 

a  =  the  base  of  the  system  of  logarithms,  and 
m  and  n  constants  to  be  determined  ; 

then  the  form  of  the  expression  for  the  law  of  pressures 
will  be 

loga  —  =  nh  j 
m 

which  may  be  written 

p  =.  ma"11. 

Thus  far  the  distance  has  been  reckoned  downward.  If 
it  be  reckoned  above  the  initial  point,  the  sign  of  h  will  be 
changed,  and  we  have 

p  =  mar1*. 

In  this  equation,  if  h  =  0,  the  pressure  will  be  represented 
by  po ;  hence, 

and  the  equation  becomes 

Let  b0  be  the  reading  of  the  barometer  at  the  initial 
station  and  b  the  reading  at  the  second  station,  then 

b      p 

and  the  preceding  equation  becomes 

b  =  b0a~nh  ; 
hence,  taking  the  logarithms  of  both  sides,  we  find 


[4JM.J  PROBLEMS. 


in  which  the  subscript,  a,  indicates  that  the  base  of  this 
system  is  a.  To  find  n  and  a  requires  at  least  two  obser- 
vations at  two  known  elevations  ;  *  but,  without  attempting 
to  find  them  in  this  place,  we  observe  that  it  has  been 
found  that  the  base  is  that  of  the  Naperian  system  of 
logarithms,  and 


n 


*  These  relations  are  easily  deduced  in  the  following  manner  :  Let  p 
be  the  pressure  due  to  one  atmosphere,  whose  height  is  H  and  density 
J>,  and  dp  the  pressure  due  to  a  lamina  whose  thickness  is  dx.  Then 


and 

dp  =  Ddx. 

Dividing  the  latter  by  the  former  gives 

dp  _  dx 

7~^; 

integrating  which  between  the  limits  of  p0  and  p,  and  0  and  a?,  gives 


328  GASES    AND    VAPORS. 

the  height  of  a  homogeneous  atmosphere,  which,  according 
to  Article  413,  is 

H=  26,170  feet  nearly. 

If  common  logarithms  are  used,  they  must  be  divided  by 
the  modulus  of  the  common  system  ;  or,  what  is  the  same 
thing,  multiplied  by  the  reciprocal  of  the  modulus,  to  re- 
duce the  result  to  an  equivalent  value.  Hence,  the  value 
of  h  becomes 

h  =  2.30258  x  26170  log  ^ 

3  0 

=  60258  log  ^. 

By  means  of  actual  observations  it  has  been  found  that  the 
coefficient  should  be  somewhat  larger  than  that  given 
above,  and  that 

h  =  60346  log  -|° 

gives  better  results. 

It  is  necessary,  however,  to  add  several  corrections  to 
this  formula.  The  pressure  of  the  air  and  the  weight  of 
the  mercury  will  both  vary  with  the  force  of  gravity.  The 
force  of  gravity  at  any  latitude,  Z,  that  at  45  degrees 
being  unity,  will  be  (see  page  32), 

0.08238 


=  1-0.00256  cos  2  Z, 

and   the  coefficient   60346    must  be  multiplied  by  this 
quantity. 

The  density  of  the  atmosphere  also  changes  with  its 
temperature.     Let  tv  be  the  temperature  at  the  first  sta- 


F424.]  PROBLEMS.  329 

tion  and  tz  that  at  the  second,  then  will  the  mean  temper- 
ature be 

iOW2), 

which  may  be  considered  as  the  temperature  above  or  be- 
low 32°.  Hence,  the  expansion  of  the  air  will  be  repre- 
sented by  the  expression 


or,  for  the  Centigrade  scale, 
l  +  0.0036 

Hence,  the  formula  for  the  height  becomes 
h  (in  feet)  =  60346(1-0.00256  cos  2i)[l+0.00102(«1+<;i-640)]log|-0  , 

which  is  the  formula  given  by  Laplace.  But  Poisson,  in 
his  Traite  de  Mecanique,  pages  622-636,  introduces  cor- 
rections for  the  expansion  or  contraction  of  the  mercurial 
column  due  to  changes  of  temperature,  as  determined  by 
an  attached  thermometer.  (Mercury  expands  0.0001001 
of  its  length  for  each  degree  F.  of  increase  of  its  tempera- 
ture). He  also  gives  corrections  for  the  diminution  of 
gravity  due  to  an  elevation.  (It  varies  inversely  as  the 
square  of  the  distance  measured  from  the  centre  of  the 
earth.)  Also  a  correction  due  to  the  attraction  of  a  moun- 
tain when  observations  are  taken  near  it.  It  is  rare,  how- 
ever, that  all  these  refinements  are  used  in  practice. 

At  the  level  of  the  sea  the  mercury  stands  at  30  inches  nearly. 

5,000  feet  above  "  "  24.7        " 

10,000  feet  (height  of  Me.  ^tna)      u  20.5        " 

15,000  feet  (height  of  Mt.  Blanc)      "  16.9        u 

3  miles  .......     16.4        " 

6  miles  .        .  8.9        «• 


330  GASES    AND    VAPORS.  ^434.] 

6.  To  find  the  velocity  of  discharge  of  ait  into  a 
vacuum  through  a  small  orifice  in  a  vessel,  the  pressure 
within  the  vessel  being  equal  to  one  atmosphere. 

This  may  be  solved  according  to  three  different  hypo- 
theses : 

1st.  Suppose  that  the  gas  is  incompressible.  In  this 
case  the  head  due  to  the  velocity  will  equal  the  height  of 
a  homogeneous  atmosphere,  and  we  have 


v  =     %gh  =G4$x  26170 
=  1300  ft.  per  sec.  nearly, 
=  886  miles  per  hour  nearly. 

2d.  Suppose  that  the  gas  is  compressible  and  perfectly 
elastic.  Then  it  may  be  shown  by  higher  analysis  that 

»=\/2Plog|-, 

in  which  P  denotes  the  pressure  of  one  atmosphere,^?  the 
pressure  within  the  vessel,  and  pl  the  external  pressure. 
But  when  the  issue  is  into  a  vacuum  p±  =  o,  and  we  have 

v  =  oo  , 

that  is,  for  this  case,  according  to  this  hypothesis,  the  velo- 
city will  be  infinite,  a  result  which  is  incorrect,  as  shown 
in  the  following  Article. 

3d.  Consider  the  gas  not  only  as  elastic  and  compressi- 
ble, but  also  that  its  temperature  and  density  cJiange  sud- 
denly as  the  gas  escapes.  It  may  be  found  for  this  case 
(see  Eankine's  Applied  Mechanics,  page  582)  that 


.p 

in  which 


']• 


[424.]  PROBLEMS.  331 

7  =  1.408. 

T0  =  493.2°  F.  above  absolute  zero  =  the  absolute 
temperature  of  melting  ice, 

T  =  the  absolute   temperature  of  the  gas   in  the 
vessel, 

p0  =  the  height  of  the  mercurial  column  at  the  level 
of  the  sea, 

80  =  the  density  of  air  compared  with  that  of  mer- 
cury, 

p2  •=.  the  pressure  per  unit  against  the  issuing  jet, 
and 

Pi  =  the  pressure  per  unit  within  the  vessel. 
If  the  jet  issue  into  a  vacuum,  we  have 

^2=0, 

and  the  equation  becomes 


(7-1)  VoJ' 

^  =  26,214,  as  given  by  Rankine,  which  is  the  height  of 

a  homogeneous  atmosphere  for  dry  air. 
Substituting  the  several  quantities  given  above  in  the 
preceding  equation  gives 

v  =  2,41 3y  —  feet  per  second. 

To 

At  the  temperature  of  freezing,  rx  =  TO,  and  we  have 
v  =  2,413  feet  per  second. 

*• 

EXAMPLES. 

1.  Required  the  number  of  degrees  through  which  a  given 
volume  of  air  must  be  heated  so  as  to  double  ita 
volume,  the  pressure  remaining  constant. 

Am.  490. 


332  GASES   AND    VAPORS.  [424] 

2.  A  tube  30  inches  long,  closed  at  one  end  and  open  at 

the  other,  was  sunk  in  the  sea  with  the  open  end 
downward,  until  the  inclosed  air  occupied  only  one 
inch  of  the  tube  ;  what  was  its  depth  ? 

Ans.  986  feet, 

3.  A  spherical  air-bubble  having  risen  from  a  depth  of 

1,000  feet  in  water,  was  one  inch  in  diameter  when 
it  reached  the  surface ;  what  was  its  diameter  at  the 
point  where  it  started  ?  Ans.  0.32  inches. 

4.  A  balloon  whose  capacity  is  10,000  cubic  feet  is  filled 

with  hydrogen  gas ;  the  specific  gravity  of  the  gas 
being  0.0690  compared  with  air,  how  many  pounds 
of  ballast  will  just  prevent  it  from  rising  ? 

5.  Five  cubic  feet  of  air  at  32°  F.  and  pressure  of  15  Ibs. 

per  square  inch,  is  confined  in  a  vessel ;  what  will  be 
the  tension  when  heated  to  400°  F.  and  the  volume 
increased  to  5.5  cubic  feet  ? 


The  following  examples  are  selected  from  the  London 
University  Examination  Papers,  from  1862  to  1870  : 

1.  Explain  the  difficulty  of  opening  a  lock-gate  when  the  water  is  at  a 

different  level  within  and  without  the  lock ;  also,  why  the  force 
required  to  open  the  gate  is  not  proportional  directly  to  the  differ- 
ence of  level. 

2.  The  weight  of  water  is  770  times  that  of  air ;  at  what  depth  in  a 

lake  would  a  bubble  of  air  be  compressed  to  the  density  of  water, 
supposing  the  law  of  Mario tfce  to  hold  good  throughout  for  the 
compression  ? 

3.  A  body  weighs  in  air  1,000  grains,  in  water  300  grains,  and  in  another 

liquid  420  grains  ;  what  is  the  specific  gravity  of  the  latter  liquid  ? 

Ans.  .82b5. 

4.  A  mercurial  barometer  is  lowered  into  a  vessel  of  water,  so  that  the 

surface  of  the  water  is  finally  six  inches  above  the  cistern  of  the 


[424.1  EXAMPLES.  333 

barometer.  What  kind  of  change  will  take  place  in  the  reading 
of  the  column  of  the  instrument  ?  Give  a  reason  for  your  repiy. 

5.  If  a  bottle  filled  with  air  be  tightly  corked,  and  lowered  into  the 
ocean,  the  cork  will  be  forced  in  at  a  certain  depth.  Why  is  this  ? 
and  what  will  take  place  if  the  bottle  be  filled  with  water  instead 
of  air? 

6  If  a  barometer  were  carried  down  in  a  diving-bell,  what  would  take 
place  ? 

7.  A  solid,  soluble  in  water  but  not  in  alcohol,  weighs  346  grains  in  air,- 

and  210  in  alcohol ;  find  the  specific  gravity  of  the  solid,  that  of 
alcohol  beirg  0.85.  Ana.  2.1625. 

8.  A  body  whose  specific  gravity  is  3.5,  weighs  4  Ibs.  in  water.     What 

is  its  real  weight  ?  Am.  5.6  Ibs. 

9.  If  as  much  additional  air  were  forced  into  a  closed  vessel  as  it  pre- 

viously contained  when  in  communication  with  the  atmosphere, 
what  would  be  the  pressure  on  a  square  inch  of  the  internal 
surface  ? 

10.  At  what  depth  in  a  lake  is  the  pressure  of  the  water,  including  the 

atmospheric  pressure,  three  times  as  great  as  at  the  depth  of  10 
feet,  on  a  day  when  the  height  of  the  liquid  column  in  a  water- 
barometer  is  33  feet  6  inches  ?  Ans.  97  feet. 

11.  A  lump  of  beeswax,  weighing  2,895  graina,  is  stuck  on  to  a  crystal 

of  quartz  weighing  795  grains,  and  the  whole,  when  suspended  in 
water,  is  found  to  weigh  390 -grains;  find  the  specific  gravity  of 
beeswax,  that  of  quartz  being  2. 65.  Ans.  .  965. 

12.  A  barometric  tube  of  half  an  inch  internal  diameter  is  filled  in  the 

usual  way,  and  the  mercury  is  found  to  stand  at  the  height  of  30 
inches.  A  cubic  inch  of  air  having  been  allowed  to  pass  into  the 
vacuum  above  the  mercury,  the  column  is  found  to  be  depressed 
5  inches.  What  was  the  volume  of  the  original  vacuum  ? 

13.  A  bottle  holds  1,500  grains  of  water,  and  when  filled  with  alcohol  it 

weighs  1,708  grains ;  but  when  empty  it  weighs  520  grains  ;  what 
is  the  specific  gravity  of  alcohol  ?  Ans.  .792. 

14.  A  piece  of  cupric  sulphate  weighs  3  ozs.  in  vacuo,  and  1.86  ozs.  in 

oil  of  turpentine  ;  what  is  the  specific  gravity  of  cupric  sulphate, 
that  of  turpentine  being  0.88  ?  Ans.  2-fo. 

15.  If  the  height  of  the  barometer  rises  from  30  inches  to  30.25  inches, 

what  is  the  increase  of  pressure  (in  ozs. )  upon  a  square  foot  ? — the 


334  GASES    AND    VAPORS.  [434.J 

weight  of  a  cubic  foot  of  water  being  taken  to  be  1,000  ozs.,  and 
the  specific  gravity  of  mercury  13. 56.  Ans.  282.5  ozs. 

16.  A  cylindrical  vessel  standing  on  a  table  contains  water,  and  a  piece 

of  lead  of  given  size  supported  by  a  string  is  dipped  into  the 
water ;  how  will  the  pressure  on  the  base  be  affected  (1)  when  the 
vessel  is  full,  (2)  when  it  is  not  full  ?  and  in  the  second  case, 
what  is  the  amount  of  the  change  ? 

17.  A  wooden  sphere  has  a  small  hole  drilled  in  it,  and  is  placed  in 

water.  Find  its  positions  of  equilibrium  ;  and  state  which  position 
is  of  stable,  and  which  is  of  unstable  equilibrium. 

18.  The  water  above  the  empty  lock  of  a  canal  is  8  feet  higher  than  the 

base  of  the  floodgates,  which  are  4  feet  broad,  and  provided  with 
handles  10  feet  long ;  find  what  force  would  have  to  be  applied 
to  the  extremity  of  the  handle  to  force  open  a  floodgate,  without 
previously  letting  in  the  water,  assuming  a  cubic  foot  of  water  to 
weigh  1,000  ozs.  avoirdupois.  Ans.  1,600  Ibs. 

19.  A  balance  is  wholly  immersed  in  water,  and  a  body  appears  to 

weigh  1  Ib. ,  the  weights  against  which  it  is  balanced  having  the 
specific  gravity  8.5.  What  will  it  appear  to  weigh  when  balanced 
against  weights  of  the  specific  gravity  11.5  ?  j-jg 

^.— Ibs. 

20.  When  a  body  is  floating  partly  immersed  in  a  liquid,  what  effect 

will  a  fall  of  the  barometer  have  upon  the  body  ? 

21.  The  specific  gravity  of  cast  copper  is  8.79,  and  that  of  copper  wire 

is  8. 88,  What  change  of  volume  does  a  kilogramme  of  cast  copper 
undergo  in  being  drawn  out  into  wire  ? 

Ans.  1.15  cubic  centimetres. 

22.  A  cylindrical  wooden  rod  of  specific  gravity  .72  and  1  centimetre  in 

diameter  is  loaded  at  one  end  with  9.08  grammes  of  lead  (specific 
gravity  =  11.35)  ;  how  long  must  the  rod  be  in  order  that  it  may 
just  float  in  water  at  the  maximum  density  ? 

23.  State  the  relation  between  the  pressure  and  density  of  an  elastic 

fluid. 

24.  A  piece  of  cork  floats  in  a  basin  of  water,  and  the  basin  is  placed 

under  the  receiver  of  an  air-pump.  State  and  explain  the  effect 
of  pumping  out  a  portion  of  the  air  in  the  receiver. 

25.  A  wineglass  is  inverted,  and  its  rim  just  immersed  in  water.  What 

would  be  the  effect  of  placing  a  small  piece  of  ice  in  the  water  be- 
neath the  glass  ? 


[434.]  EXERCISES.  335 

26.  Find  the  atmospheric  pressure  on  a  square  inch,  assuming  that  the 

height  of  a  column  of  water  supported  by  the  atmospheric  pres- 
sure is  30  feet,  and  that  a  cubic  fathom  of  water  weighs  six  tons. 

27.  Compare  the  depths  to  which  a  right  cone  must  be  immersed  in  a 

fluid  of  twice  its  density,  that  it  may  be  in  equilibrium  when  (1) 
the  vertex  is  downwards,  and  (2)  the  base. 

88.  A  floodgate  is  6  feet  wide  and  12  feet  deep.  Reckoning  the  weight 
of  a  cubic  fathom  of  water  at  6  tons,  what  is  the  total  pressure  on 
the  floodgate  when  the  water  is  level  with  its  top  ;  and  what  is 
the  situation  of  the  centre  of  pressure  ? 

29.  A  cubic  inch  of  one  of  two  liquids  weighs  a  grains,  and  of  the  other 

b  grains.  A  body  immersed  in  the  first  fluid  weighs  p  grains,  and 
immersed  in  the  second  fluid  weighs  q  grains.  What  is  its  true 
weight,  and  what  is  its  volume  ? 

Ans.  V  --  (p-q)  +  (b-a),  W  =  (bp-aq)-s-(b-a). 

30.  A  quantity  of  air  contained  in  a  spherical  vessel  is  transferred  first 

into  a  cylindrical  vessel,  and  then  into  a  cubical  vessel,  each  of 
which  would  just  circumscribe  the  spherical  vessel.  Compare  the 
total  pressure  produced  by  the  air  on  the  walls  of  the  three  vessel* 

Ans.  5if*  :  12" :  192. 


APPENDIX. 


TABLE  I. 
EXPERIMENTS  ON  FRICTION,   WITHOUT  UNGUENTS.      BY  M.  MORIN. 


SURFACES  OF  CONTACT. 

FBICTION  OP 
MOTION. 

FRICTION  or 
QUIESCENCE. 

Coefficient 
of  Friction. 

Limiting 
Angle  of 
Resistance. 

Coefficient 
of  Friction. 

Limiting 
Angle  of 
Resistance. 

Oak  upon  oak,  the  direction  of  the  fibres 
being  parallel  to  the  motion  

0-478 

0-3534 
0-336 

0-192 

25°  33' 

17    58 
18    35 

10    52 

0-625 
0-540 

82°    V 

28    23 

Oak  upon  oak,  the  direction  of  the  fibres 
of  the  moving  surface  being  perpen- 
dicular to  those  of  the  quiescent  sur- 
face and  to  the  direction  of  the  mo- 
tion *  

Oak  upon  oak,  the  fibres  of  both  surfaces 
being  perpendicular  to  the  direction  of 
the  motion  

Oak  upon  oak,  the  fibres  of  the  moving 
surface  being  perpendicular  to  the 
surface  of  contact,  and  those  of  the 
surface  at  rest  parallel  to  the  direction 
of  the  motion  

0-271 
0-43 

15    10 
23    17 

Oak  upon  oak,  the  fibres  of  both  surfaces 
being  perpendicular  to  the  surface  of 
contact,  or  the  pieces  end  to  end  ...  . 

*  The  dimensions  of  the  surfaces  of  contact  were  in  this  experiment  '947  square  feet, 
and  the  results  were  nearly  uniform.  When  the  dimensions  were  diminished  to  -043,  • 
tearing  of  the  fibre  became  apparent  in  the  case  of  motion,  and  there  were  symptoms  of 
the  combustion  of  the  wood ;  from  these  circumstances  there  resulted  an  irregularity  in 
the  friction  indicative  of  excessive  pressure. 

15 


338  ELEMENTARY    MECHANICS. 

TABLE  I.— Continued. 


STJBFACES  OP  CONTACT. 

FBICTION  OF 
MOTION. 

FRICTION  OF 
QUIESCENCE. 

Coefficient 
of  Friction. 

Limiting 
Angle  of 
Resistance. 

Coefficient 
of  Friction. 

Limiting 
Angle  of 
Ketiistanee. 

Elm  upon  oak,  the  direction  of  the  fibres 
being  parallel  to  the  motion  

0-432 
0-246 

0-450 
0-400 

0-355 
0-36X) 
0-370 
0-400 
0-619 

0.256 
0-252 

0-194 

0-138 
0-490 

23°  22' 
13    50 

24    16 

21     49 

19    33 
19    48 
20    19 
21    49 
31    47 

14    22 
14      9 

10    59 

7    52 
26      7 

0-694 
0-376 

0-570 

0-570 

0-520 
0-53 
0-440 
0-570 
0-619 

0-649 

34°   46' 
20    37 

29    41 
29    41 

27    29 
27    56 
23    45 

29    41 

31    47 

33      0 

Oak  upon  elm   ditto  *.           

Elm  upon  oak,  the  fibres  of  the  moving 
surface  (the  elm)  being  perpendicular 
to  those  of  the  quiescent  surface  (the 
oak)  and  to  the  direction  of  the  mo- 
tion ....       .         

Ash  upon  oak,  the  fibres  of  both  surfaces 
being  parallel  to  the  direction  of  the 
motion  

Fir  upon  oak,  the  fibres  of  both  surfaces 
being  parallel  to  the  direction  of  the 
motion  ....       .       .         .           

Beach  ujion  oak,  ditto  

Wild  pear  tree  upon  oak,  ditto  

feervice  tree  upon  oak,  ditto  

Wrought  iron  upon  oak,  ditto  t  

Wrought  iron  upon  oak,  the  surfaces  be- 
ing greased  and  well  wetted.       

Wr  ought  iron  upon  elm       .       . 

Wrought  iron  upon  cast  iron,  the  fibres 
of  the  iron  being  parallel  to  the  motion 
Wrought  iron  upon  wrought  iron,    the 
fibres  of  both  surfaces  being  parallel 
to  the  motion  

0-194 
0-137 

10    59 
7    49 

Cast  iron  upon  oak   ditto         

Ditto,  the   surfaces  being  greased   and 
wetted  

0-646 

32    52 

0-195 
0-152 

0-314 
0-147 

11      3 

8    39 

17    26 
8    22 

Cast  iron  upon  cast  iron   

0-162 

9    13 

Ditto,  wator  being  interposed  between 
the  surfaces         

Cast  iron  upon  brass                                  . 

*  It  is  worthy  of  remark  that  the  friction  oak  upon  elm  is  but  five-ninths  of  that  of 
elm  upon  oak. 

t  In  the  experiments  in  which  one  ot  the  st  rfaces  was  of  metal,  small  particles  of  the 
metal  began,  after  a  time,  to  be  apparent  upon  the  wood,  giving  it  a  polished  metallic 
appearance ;  these  were  at  every  experiment  wiped  off ;  they  indicated  a  wearing  of  th« 
metal.  The  friction  of  motion  and  that  of  quiescence,  in  these  experiments,  coincided 
The  results  were  remarkably  uniform. 


APPENDIX. 
TABLE  I.— Continued. 


SURFACES  OF   CONTACT. 


Oak  upon  cast  iron,  the  fibres  of  the 
wood  being  perpendicular  to  the  direc- 
tion of  the  motion 

Hornbeam  upon  cast  iron — fibres  parallel 
to  motion 

Wild  pear  tree  upon  cast  iron — fibres 
parallel  to  the  motion 

Steel  upon  cast  iron 

Steel  upon  brass 

Yellow  copper  upon  cast  iron 

Do.  do.  oak 

Brass  upon  cast  iron 

Brass  upon  wrought  iron,  the  fibres  of 
the  iron  being  parallel  to  the  motion. . 

Wrought  iron  upon  brass 

Brass  upon  brass 

Black  leather  (curried)  upon  oak  * 

Ox  hide  (such  as  that  used  for  soles  and 
for  the  stuffing  of  pistons)  upon  oak, 
rough 

Do.  do.  smooth 

Leather  as  above,  polished  and  hardened 
by  hammering 

Hempen  girth,  or  pulley-band  (sangle  de 
chanvre),  upon  oak,  the  fibres  of  the 
wood  and  the  direction  of  the  cord  be- 
ing parallel  to  the  motion 

Hempen  matting,  woven  with  small 
cords,  ditto 

Old  cordage,  1 3^  inch  in  diameter,  ditto 

Calcareous  oolitic  stone,  used  in  build- 
ing, of  a  moderately  hard  quality, 
called  stone  of  Jaumont— upon  the 
same  stone 

Hard  calcareous  stone  of  Brouck,  of  a 
light  gray  color,  susceptible  of  taking 
a  tine  polish  (the  muschelkalk),  mov- 
ing upon  the  same  stone 


FKICTION  OP 
MOTION. 

FRICTION  OK 

QUIESCENCE. 

Coefficient 
of  Friction. 

Limiting 
Angle  of 
Resistance. 

Coefficient 
•  of  Friction. 

Limiting 
Angle  of 
Resistance. 

0-372 
0-394 

0-436 
0-202 
0-152 
0-189 
0-617 
0-217 

0-161 
0-172 
0-201 
0-265 

0-52 

o-ass 

0-296 

0-52 

0-32 
0-52 

0-64 
0-38 

20°   25> 
21     31 

23    34 
11     26 
8    39 
10    49 
31    41 
12    15 

9      9 
9    46 
11    22 
14    51 

27    29 
18    31 

16    30 

27    29 

17    45 

27    29 

32    38 
20    49 

0-617 

81    41 

0-74 

0-605 
0-43 

36    31 

31    11 
23    17 

0-64 

0-50 
0-79 

0-74 
0-70 

32    38 

2t    84 
38    19 

36    3l 
35      t 

*  The  friction  of  motion  was  very  nearly  the  same  whether  the  surface  of  contact  WM 
inside  or  the  outside  of  the  skin.  The  constancy  of  the  coefficient  of  the  friction  of 
motion  was  equally  apparent  in  the  rough  and  the  smooth  skins. 


340  ELEMENTARY    MECHANICS. 

TABLE  I.— Continued. 


FRICTION  OF 

FBICTION  OF 

MOTION. 

QUIESCENCE. 

SUBFAOES  OP  CONTACT. 

•M  a 
c  g 

**i 

5*Hc*x 

.2  c  o 

£  0 

<**-! 

11 

•SB 

.5  c§ 

g 

StK'S 
.__   C    £ 

O  o 

00 

^><K 

The  soft  stone  mentioned  above,  upon 

the  hard  

0'65 

33° 

y 

0-75 

36°  53' 

The  hard   stone  mentioned  above  upon 

the  soft  

0-67 

33 

50 

0-75 

36    53 

Common  brick  upon  the  stone  of  Jau- 

mont    .   .       

0-65 

33 

fl 

0-65 

33      2 

Oak  upon  ditto,  the  fibres  of  the  wood 

being  perpendicular  to  the  surface  of 

the  stone  

0-38 

90 

4°- 

0-63 

32    13 

Wrought  iron  upon  ditto  

0-69 

34 

S7 

0-49 

26      7 

Common  brick  upon  the  stone  of  Brouck 
Oak  as  before  (endwise)  upon  ditto  

0-60 
0-38 

30 
20 

58 
49 

0-67 
0-64 

33    50 
32    38 

Iron,                    ditto           ditto       .... 

0-24 

18 

30 

0-48 

22    47 

APPENDIX. 


341 


TABLE   II. 


EXPERIMENTS  ON  TOE  FBICTION  OF  UNCTUOUS  SUBFACES. 
BY  M.  MORIN. 

IQ  these  experiments  the  surfaces,  after  having  been  smeared  with  an 
nuguent,  were  wiped,  so  that  no  interposing  layer  of  the  unguent  prevented 
their  intimate  contact. 


SURFACES  OF  CONTACT. 

FBICTION  or 
MOTION. 

FRICTION  o» 
QUIESCENCE. 

Coefficient 
of  Friction. 

Limiting 
Angle  of 
Resistance. 

Coefficient 
of  Friction. 

!^  £f  Limiting 
Angle  of 
g  5  Resistance. 

Oak  upon  oak,  the  fibres  being  parallel 

to  t.h"  motion    ... 

0-108 

0-143 
0-136 
0-119 
0-330 
0-140 
0-133 
0-177 

6"   107 

8      9 
7    45 
6    48 
18    16 
7    59 
7    52 
10      3 

0-390 
0-314 

Ditto,  the  fibres  of  the  moving  body  be- 
ing perpendicular  to  the  motion  

Oak  upon  elm   fibres  parallel.       ... 

Rim  upon  oak,  ditto  . 

0-420 

22    47 

Beech  upon  oak  ditto  

Elm  upon  elm   ditto  

Wrought  iron  upon  elm   ditto  

Ditto  upon  wrought  iron  ditto.  .   ...... 

0-118 

6    44 

Cast  iron  upon  wrought  iron,  ditto  
Wrought  iron  upon  brass  

0-143 
0-160 
0-166 
0-107 

0-125 

0-137 
0-135 
0-144 
0-132 
0-107 
0-134 

o-ioo 

0-115 

0-229 
0-244 

8      9 
9      6 
9    26 
6      7 

7      8 

7    49 
7    42 
8    12 
7    32 
6      7 
7    38 
5    43 
6    34 

13    54 
13    43 

Brass  upon  wrought  iron  

Cast  iron  upon  oak,  ditto  

o-ioo 

5    43 

Ditto  upon  elm.  ditto,  the  unguent  being 
tallow                   

Ditto,   ditto,    the  unguent  being  hog's 
lard  and  black  lead   

6:098 

*5"36 

Kim  upon  cast  iron,  fibres  parallel  

0-164 

9    19 

Leather  (ox  hide)  well  tanned  upon  cast 

0-267 

14    57 

ELEMENTARY    MECHANICS. 


TABLE  in. 

EXPERIMENTS  ON  FKICTION  WITH  UNGUENTS  INTERPOSED. 
BY  M.   MOKIN. 

The  extent  of  the  surfaces  in  these  experiments  bore  such  a  relation  to  th< 
pressure  as  to  cause  them  to  be  separated  from  one  another  throughout  by  at 
interposed  stratum  of  the  unguent 


FBICTION  OF 
MOTION. 

FBICTION  OF 
QUIESCENCE. 

Coefficient  of 
Friction. 

Coefficient  of 
Friction. 

Oak  upon  oak,  fibres  parallel  

0'164 

0-440 

Dry  soap. 

Ditto        ditto         

0-075 

0-164 

Tallow. 

Ditto        ditto          

0-067 

Hog's  lard. 

Ditto,  fibres  perpendicular  

0-083 

0'254 

Tallow. 

Ditto        ditto          

0-072 

Hog's  lard. 

Ditto        ditto          

0-250 

Water. 

Ditto  upon  elm,  fibres  parallel  .  . 

0-136 

Dry  soap. 

Ditto        ditto          

0-073 

0-178 

Tallow. 

Ditto        ditto          

0-066 

Hog's  lard. 

0-080 

Tallow. 

Ditto  upon  wrought  iron,  ditto.. 

0-098 

Tallow. 

Beech  upon  oak.  ditto  . 

0-055 

Tallow. 

Kim  upon  oak,  ditto  

0-137 

0-411 

Dry  soap. 

Ditto        ditto         

0-070 

0-142 

Tallow. 

Ditto        ditto          

0-060 

Hog's  lard. 

Ditto  upon  elm,  ditto  

0-139 

0-217 

Dry  soap. 

Ditto  upon  cast  iron,  ditto  

0-066 

Tallow. 

Wrought  iron  npon  oak,  ditto  
Ditto        ditto        ditto      

0-256 
0-214 

0-649 

(  Greased  anc 
•<  saturated 
(  with  water 
Dry  soap. 

Ditto        ditto        ditto      

0-085 

0-108 

Tallow. 

Ditto  upon  elm,     ditto       

0-078 

Tallow. 

Ditto        ditto        ditto      

0-076 

Hog's  lard. 

Ditto        ditto        ditto       

0-055 

Olive  oil 

Ditto  upon  "ast  iron,  ditto  

0-103 

Tallow. 

Ditto        ditto        ditto       

0-076 

Hog's  lard. 

Ditto        ditto        ditto       

0-066 

0-100 

Olive  oil. 

Ditto  upon  wrought  iron,  ditto.  . 
Ditto        ditto        ditto       

0-082 
0-081 

Tallow. 
Hog's  lard. 

Ditto        ditto        ditto       

0-070 

0-115 

Olive  oil. 

Wrought  iron  upon  brass,  fibres  | 
parallel  J 

0-103 

Tallow. 

0-075 

Hog's  lard. 

Ditto        ditto        ditto      

0-078 

Olive  oil. 

0-189 

Drv  soao. 

APPENDIX. 
TABLE  III.— Continued. 


343 


FBICTION  OF 
MOTION. 

FBICTION  OF 
QUIESCENCE. 

Coefficient  of 
Friction. 

Coefficient  of 
Friction. 

Cast  iron  upon  oak,  fibres  parallel. 
Ditto        ditto        ditto       

0-218 
0-078 

0-646 

o-ioo 

(  Greased,  and 
•<  saturated 
(  with  water. 
Tallow. 

Ditto        ditto        ditto       

0-075 

Hog's  lard. 

Ditto        ditto        ditto       

0-075 

o-ioo 

Olive  oil. 

Ditto  upon  elm       ditto         

0-077 

Tallow 

Ditto        ditto        ditto       

0-061 

Olive  oil. 

Ditto        ditto        ditto       

0-091 

(  Hog's  lard  & 

Ditto  upon  wrought  iron  

o-ioo 

j  plumbago. 
Tallow. 

Cast  iron  upon  cast  iron  

0-314 

Water. 

Ditto        ditto         

0-197 

Soap. 

Ditto        ditto         

o-ioo 

o-ioo 

Tallow. 

Ditto        ditto        

0-070 

o-ioo 

Hog's  lard. 

Ditto        ditto        

0-064 

Olive  oil. 

Ditto        ditto         

0-055 

j  Lard  and 

Ditto  upon  brass  

0-103 

(  plumbago. 
Tallow. 

Ditto        ditto        

0-075 

Hog's  lard. 

Ditto        ditto        

0-078 

Olive  oil. 

Copper  npon  oak,  fibres  parallel.  .  . 
Yellow  copper  upon  cast  iron  

0-069 
0-072 

o-ioo 

0-103 

Tallow. 
Tallow. 

Ditto        ditto         

0-068 

Hog's  lard. 

Ditto        ditto        

0-066 

Olive  oiL 

Brass  upon  cast  iron  

0-08ft 

0-106 

Tallow. 

Ditto        ditto        

0-077 

Olive  oiL 

Ditto  upon  wrought  iron  

0-081 

Tallow. 

Ditto        ditto         

0-089 

t  Lard  and 

Ditto        ditto        

0-072 

"(  plumbago. 
Olive  oil. 

0-058 

Olive  oiL 

Steel  upon  cast  iron  

0-105 

0-108 

Tallow. 

Ditto        ditto               

0-081 

Hog's  lard. 

Ditto        ditto        

0-079 

Olive  oil. 

0-093 

Tallow. 

Ditto        ditto                     

0-076 

Hog's  lard. 

0-056 

Tallow. 

Ditto        ditto                       

0-053 

Olive  oil 

Ditto        ditto        

0-067 

,  Lard  and 

Tanned  ox  hide  upon  cast  iron  
Ditto        ditto        

0-365 
0-159 

i  Greased,  and 
-<  saturated 
(  with  water. 
Tallow. 

Ditto        ditto        

0-133 

0.122 

Olive  oiL 

0-241 

Tallow. 

344 


ELEMENTARY    MECHANICS. 
TABLE  III.— Continued. 


FBICTION  OF 
MOTION. 

FnicnoN  OF 
QUIESCENCE. 

Coefficient  of 
Friction. 

Coefficient  of 
Friction. 

Tanned  ox  hide  upon  brass     

0-191 

Olive  oiL 

Ditto  upon  oak  

0-29 

0-79 

Water. 

Hempen  fibres  npt  twisted,  mov-' 
ing  upon  oak,  the  fibres  of  the 
hemp  being  placed  in  a  direc- 
tion perpendicular  to  the  direc- 
tion of  the  motion,  and  those  of 
the  oak  parallel  to  it  

0-332 

0-869 

(  Greased,  and 
•<  saturated 
(  with  water. 

The  same  as  above,  moving  upon 
cast  iron  

0-194 

Tallow. 

Ditto  

0-153 

Olive  oil. 

Soft  calcareous  stone  of  Janmont 
upon  the  same,  with  a  layer  of 
mortar,  of  sana,  and  lime  inter- 
posed, after  from  10  to  15  min- 
utes' contact  

0-74 

TABLE  IV. 
OP  THE  SPECIFIC   GRAVITIES  OF  BODIES. 

[The  density  of  distilled  water  is  reckoned  in  this  Table  at  its  maximum, 
38%«  P.  =  1.000.] 


Name  of  the  Body. 


Specific  Gravity. 


L  SOLID  BODIES. 

(1)  METALS. 


Antimony  (of  the  laboratory) 

Brass 

Bronze  for  cannon,  according  to  Lieut.  Matzka . 

Ditto,  mean 

Copper,  melted. 

Ditto,  hammered 

Ditto,  wire-drawn 

Gold,  melted 

Ditto,  hammered 

Iron,  wrought 


4-2 

7-6 

S-414 
8-758 
7-788 
8-878 
8-78 
10*888 
19-::iil 
7-207 


APPENDIX. 
TABLE  TV  .—Continued. 


348 


Name  of  the  Body. 


Specific  Gravity. 


I.  SOLID  BODIES. 

Iron,  cast,  a  mean 7*251 

Ditto,  gray 7*2 

Ditto,  white 7*5 

Ditto  for  cannon,  a  mean 7*21 

Lead,  pure  melted 11*3303 

Ditto,  flattened 11*388 

Platinum,  native 16*0 

Ditto,  melted 20*855 

Ditto,  hammered  and  wire-drawn 21  '25 

Quicksilver,  at  82°  Fahr 13*568 

Silver,  pure  melted 10-474 

Ditto,  hammered 10 '51 

Steel,  cast 7*919 

Ditto,  wrought 7 '840 

Ditto,  much  hardened 7'818 

Ditto,  slightly 7*833 

Tin,  chemically  pure 7*291 

Ditto,  hammered. 7' 299 

Ditto,  Bohemian  and  Saxon 7*312 

Ditto,  English 7*291 

Zinc,  melted 6*861 

Ditto,  rolled 7*191 

(2)  BUILDING  STONES. 

Alabaster 2*7 

Basalt 2*8 

Dolerite 2*72 

Gneiss 2*5 

Granite 2*5 

Hornblende 2*9 

Limestone,  various  kinds 2*64 

Phonolite 2*51 

Porphyry 2*4 

Quartz 2*56 

Sandstone,  various  kinds,  a  mean 2*2 

Stones  for  building 1  *66 

Syenite 2*5 

Trachyte 2*4 

Brick 1*41 

(3)  WOODS.  Fresh  felled.      Dry. 

Alder  .. 0-8571        0'5001 

Ash  0-9086       0*0440 

Aspen.'.'.       0-7654   0*4303 

Birch..  0*9012   0*6274 

Bos  0*9822        0*5907 

Elm"  0*9476   0-5474 

Fir"  , 0*8941   0*5550 

Hornbeam.'.     0*9452   0*7695 

Horse-chestnut 0*8614        0*5749 


15* 


346 


ELEMENTARY    MECHANICS 
TABLE  IV.  —  Continued. 


Name  of  the  Body. 

Specific  Gravity. 

I.  SOLID  BODIES. 
Larch.  

Fresh  felled.       Dry. 
0'9206        0-4735 

Lime  

0'8170        O'431'O 

Maple  

0-903(5       ()-i;:'.x! 

Oak    

1-0494        0-<;777 

Ditto,  another  specimen.  

1-0754        0'7075 

Pine,  Pimi*  Abies  Picea  

0*8699        0-4716 

Ditto,  Plnus  Sylvestris  

0-9121        0-5503 

T*oplar  (Italian)     

0*7634        0-3931 

Willow  

0-7155        0-5389 

Ditto,  white.  

0-9859        0-4873 

(4)  VAKIOTJS  SOLID  BODIES. 
Charcoal,  of  cork  ....           

O'l 

Ditto  soft  wood.           .                  ..         .         . 

0-38 

Ditto,   oak  

1-573 

Coal  

1-233 

Coke  

1*865 

Earth,  common      

1-48 

rough  sand  

1-92 

rotifh  earth,  with  gravel  

2-02 

moist  sand  

2-05 

gravelly  soil  

2-07 

clay  

2-15 

clay  or  loam,  with  gravel  

2'48 

Flint,  dark.  

2-542 

Ditto,  white   

2  '741 

Gunpowder,  loosely  filled  in, 
coarse  powder   

0*886 

musket  ditto  

0-993 

Ditto,  slightly  shaken  down, 
musket  powder  ..       ..         ,  

1-069 

Ditto    solid  

2-248 

Ice  

0-916 

Lime   unslacked  

1-842 

Ill  sin.  common.  

1-089 

2-257 

Saltpetre,  melted.  

2-745 

Ditto,  cr  ystallized  

1-900 

Slate-pencil  

1-8 

Sulphur.  

1'92 

Tallow  

0-942 

Turpentine.       .   ..         

0-991 

Wax,  white  

0-969 

Ditto,  yellow  

0*965 

Ditto,  shoemaker's  

0-897 

n.  LIQUIDS. 

Acid,  acetic  

1-063 

1-211 

Ditto,  nitric,  concentrated  

1-521 

APPENDIX. 
TABLE  IV.-^-Uontinued. 


347 


Name  of  the  Body. 


Specific  Gravity. 


Acid,  sulphuric,  English 1  '845 

Ditto,  concentrated   (Nordh.) 1  -860 

Alcohol,  free  from  water 0'792 

Ditto,  common 0-824 

Ammoniac,  liquid 0'875 

Aquafortis,  double '. 1  -300 

Ditto,  single 1  -200 

Beer 1  •  023 

Ether,  acetic 0'866 

Ditto,  muriatic 0*845 

Ditto,  nitric 0'886 

Ditto,  sulphuric 0* 715 

Oil,  linseed. 0'928 

Ditto,  olive 0'915 

Ditto,  turpentine 0'793 

Ditto,  whale 0'923 

Quicksilver 13-568 

Water,  distilled 1  -000 

Ditto,  rain 1-0013 

Ditto,  sea 1'0265 

Wine 0-992 

III.    GASES.  Barometer  80  In. 

Temperature  —  32° 

Atmospheric  air  =  -rhr  = I'OOOO 

Carbonic  acid  gaa 1  '5240 

Carbonic  oxide  gas 0"9569 

Carburetted  hydrogen,  a  maximum 0'9784 

Ditto  from  coals jjigg 

Chlorine 2 '  4700 

Hydriodic  gas 4*4430 

Hydrogen 0-0688 

Hydrosulphuric  acid  gas 1*1912 

Muriatic  acid  gas 1  '2474 

Nitrogen. 0'9760 

Oxygon 1-1026 

Phosphuretted  hydrogen  gas 0'8700 

Steam  at  212°  Pahr 0'6235 

Sulphurous  acid  gas 2'2470 

I 


ANSWERS. 


Pages  13,  14.     1.  250  ft.  and  50  ft.     2.  25  ft.  and  12$  ft.     3.  112  ft. 
4.  6*  ft.     5.  |V2.     6.  0-5466 +  .     7.9-8. 

Pages  35,  36.     1.    80 "2  ft.     ii.    9 '3  sec.     3.  155-4  ft.    4.   1930  ft. 


5.  39-23  m.    6.  91 -35  ft.  7.   2'35sec.    1O. 


Page  47.     1.  35-87  ft.     2.   80-,'V  ft.     3.  13-88  +  ft.     4.  31*  Ibs. 

Page  50.  1.  201^-  ft.  2.  134-^  ft.  3.  52  '55  Ibs.  and  47'57  Ibs. 
4.  64-6  ft.  5.  i§  Ibs.  ;  3*  Ibs.  ;  10  Ibs.  ;  oo.  6.  2-6+  sec. 

Pages   64,   65.      1.   3168   cu.   ft.     2.    197,920'8   ft.  -Ibs.     3.   7£  in. 

4.  7-217  +  .     5.   29,057.     6.   37'87+.      7.    1-6  ft.     §.    IJs  H>. 
9.  0-1668  +  .     1O.  16-75.  ft. 

Pages  75,  76.   1.  1  -07+  ft.  -Ibs.    2.18ft.   3.  2'53  miles.    4.  19'2-lbs. 

5.  3Mt.     6.  160  -8  ft.  perm.  fV~~&t       I  [Q  *T 
Pages  91,  92.     1.  25  ibs.—  sec.     2.  0  '05538  inches.    3.  24,446,199  Ibs. 

Page  99.  7.  Impossible,  unless  the  angle  35°  be  considered  as  the  sup- 
plement of  the  real  angle.  If  so  considered,  then  angle  (RF)=30°. 
^=90*63  Ibs.  and  #=57-34  Ibs.  §.  12  Ibs.  ;  16  Ibs. 

Page  103.     2.  22  "36. 

Pages  115,  116.     I.  144"2  Ibs.     2.  120  Iba     3.  60  Ibs.     4.    V2AC. 

BD 
5.  85-97+  Ibs.  ;  42-98+  Ibs.     6.  inBACW-     7*  Bisect  the 


angle.     8.  931  -8  Ibs.     9.  oo. 
Page  133      3.  14  Ibs.  ;  3f  feet.     4.  18  Ibs.  ;  4  feet  3i  inches. 

PF 
5. 


p  , 


350  ELEMENTARY  MECHANICS. 

fan  0 

}>'-j-x  area  A  GBC-  \  CH.  ABC 
Page  154.      2.  -j&iAHBff  '  '  9.  **V0>. 

4.  fr(ABf.     5.  \*(ABF\ 
Page  169.     1.  48'6  ft.     2.  (Make  v  =  0  in  Eq.  1,  p.  165)  PsinB  = 

WsiaA.    3.  P=  1-38  W.     4.  8°  37'  40".     6.  #Cfr=sin-'Q,  ^    v- 

o(.n—  r) 


Pages  179,  180.  1.  28  -8  Ibs.  2.  60  Ibs.  3.  Wain  A.  4.  WtanA. 
5.  161  Ibs.  ;  34-4  Ibs.  6.  60  Ibs.  7.  9°  35'  40".  §.  28°  57'  17". 
9.  P=W;  60°.  1O.  *'3feet. 

Pages  192,  193.  1.90  -4  Ibs.  3.  TFcotfl;  WcosecB.  4.  t  =  250  Ibs.  , 
c  =  354  Ibs.  5.30°.  6.  26°  33'  54". 

Page  201.  2.  b  =  3'10  in.  ;  d  =  12  43  in.  3.  9  29  in.  4.  11111  Ibs. 
5.  24-5  in.  6.  76,800  Ibs.  7.  23  ft.  6  -8  in. 

Pages  208,  209.  1.  14°  24'.  2.  47'65  ft.  ;  -34  "3  ft.  ;  -  637'3  ft. 
3.  25  '36  ft.  4.  Draw  a  line  from  the  point  to  the  extremity  of  the 
vertical  diameter  of  the  circle,  and  the  path  required  will  be  the  ex- 
ternal part  of  the  secant.  5.  26-0.")  miles  per  hour.  6.  4857  '.li  ft. 

7.  IGl^sec.  ;  302  '8  sec.     §.  32'34'ft.     9.  1903'64ft. 

tfl.r 
Pages  217,  218.    1.  8046Lft  ;  201  ft.    2.  ^6  -8  ft.    5.  45.    6.40?r67ft. 

7.  17-6  ft.  §.  228-2  ft.  per  sec.  (velocity  of  projection)  ;  15°  49  9"  ; 
850ft  ;  3  -9  sec. 

Pages  233,  234.     1.  v  =  *2ffi:     2.38-3.     3.  38  "3.     4.  2  W. 

5     -®*/9        6.    «k£9  IP.      7.    61-14  m.  per  h.     §.1-12  in. 
'    v  V  nR 

9.  1-93  in.  1O.  76°  25'  40"  ;  @C  =  14  -58  in.  ;&C-  3  '52  inches. 
11.  118'95.  12.  Let  8  =  the  weight  of  a  unit  of  volume,  then  the 

13786-7 

number  of  revolutions  per  minute  will  be  --  =  —  .     If  the  material 

V5 

weighs  144  Ibs.  per  cubic  foot  (which  i.s  somewhat  too  small  for 
most  stone,  but  sufficiently  exact  for  a  near  approximation),  we 
have  1148-9  turns  per  minute,  or  19'15  turns  per  second. 

Pages  248,  249.     3.  0  '0181  in.    4.  11024-3  ft    5.  3-9168  sec.    6.4-9 


ANSWERS.  351 

miles  per  second.    7.  42  m.  14  sec.    8.  0'453  ft.  per  sec.    9.  0'0148 
second. 

Pages  258,  259.  1.  Pressure  on  the  bottom  and  sides,  58670 '8  +  Iba 
2.  1-730  Ibs.  3.  195-96  Ibs.  '  <l.  28  Ibs.  5.  50'73  ft.  per  sec. 
6.  396  -56  Ibs  7.  Let  a  be  the  half  length  of  the  bar,  s  the  specific 
gravity  of  the  wood  in  reference  to  the  liquid,  and  x  the  distance 
from  the  point  of  attachment  of  the  cord  to  the  middle  point  of 

Is 

the  bar  ;  then  x  = a.     8.  81  inches. 

18s— 4 

Page  270.     2.  4i.     3.  240  grs.     4.  1796875  Ibs.     5.  J.     6.  1'6. 

Page  276.  1.  16°  41'  57".  2.  The  box  being  so  deep  that  none  of  the 
liquid  will  flow  over,  we  have  21  $•  ft.  per  sec.  3.  g.  4.  54*16. 

Page  282.  2.  82'28  Ibs.  ;  164-57  Ibs.  3.  1767"!  Ibs. ;  589-0  Ibs.  ; 
589-0  Ibs.  4.  98,175  Ibs.  ;  32,725  Ibs.  5.  4312£  Ibs.  6.  2250  Iba. 

Pages  287,  288.    2.  5 19  in.    3.  2  ft.'  8}  in.    4.  6 "2+  ft.    5.  50,000  Ibs. 

Page  310.     1.  6  m.  41-2  sec.      2.  13618-8  cu.  ft.      3.  x  —  -y4;  0'134 

27 

sq.  inches.     4.  197.+  cu.  ft. 
Page  332.     4.  751 '34  Ibs.     5.     23 -868  Ibs. 


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